:: Solutions of Linear Equations
:: by Karol P\c{a}k
::
:: Received December 18, 2007
:: Copyright (c) 2007-2016 Association of Mizar Users
:: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.
environ
vocabularies NUMBERS, NAT_1, VECTSP_1, FREEALG, FINSET_1, SUBSET_1, MATRIX_1,
TREES_1, FINSEQ_1, RELAT_1, XXREAL_0, INCSP_1, FVSUM_1, CARD_3, RVSUM_1,
GROUP_1, ARYTM_1, SUPINF_2, FUNCT_1, ARYTM_3, CARD_1, XBOOLE_0, ALGSTR_0,
ORDINAL4, STRUCT_0, FINSEQ_2, TARSKI, RLVECT_1, PARTFUN1, VALUED_1,
ZFMISC_1, CLASSES1, XCMPLX_0, MATRIX_3, MATRIX13, LAPLACE, PRVECT_1,
RLVECT_3, LMOD_7, QC_LANG1, RLSUB_1, MESFUNC1, PRE_POLY, MATRIX11,
MATRIXR1, AFINSQ_1, FINSEQ_3, FUNCT_2, SETWISEO, FINSOP_1, RLVECT_2,
RLVECT_5, FUNCOP_1, FUNCT_4, MATRIX_6, MATRIX15;
notations TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, ORDINAL1, NUMBERS, CARD_1,
XCMPLX_0, ALGSTR_0, XXREAL_0, NAT_1, FINSET_1, RELAT_1, FUNCT_1,
PARTFUN1, FUNCT_2, SETWOP_2, FINSEQ_1, STRUCT_0, FUNCOP_1, RLVECT_1,
GROUP_1, VECTSP_1, FINSEQ_2, MATRIX_0, MATRIX_1, FVSUM_1, FINSOP_1,
NAT_D, GROUP_4, MATRIX_3, MATRIX_6, MATRIX_7, FINSEQ_7, FUNCT_4,
DOMAIN_1, MATRIX11, MEASURE6, VECTSP_4, VECTSP_6, VECTSP_7, VECTSP_9,
PRVECT_1, MATRIX13, PRE_POLY, MATRLIN, FUNCT_7, PNPROC_1, LAPLACE;
constructors SETWISEO, FINSOP_1, ALGSTR_1, FVSUM_1, GROUP_4, FINSEQ_7,
WELLORD2, VECTSP_6, VECTSP_7, VECTSP_9, MATRIX_6, MATRIX_7, MATRIX11,
MATRIX13, PNPROC_1, REALSET1, LAPLACE, BINARITH, RELSET_1;
registrations FUNCT_1, FINSET_1, STRUCT_0, FVSUM_1, VECTSP_1, FUNCT_2,
ORDINAL1, XXREAL_0, NAT_1, FINSEQ_1, RELAT_1, CARD_1, FUNCOP_1, SETFAM_1,
FINSEQ_2, MATRLIN, MATRIX13, XREAL_0, VECTSP_9, VALUED_0, RELSET_1,
PRVECT_1, MATRIX_0, MATRIX_1;
requirements ARITHM, NUMERALS, REAL, BOOLE, SUBSET;
definitions TARSKI, XBOOLE_0, MATRIX_1;
equalities XBOOLE_0, FINSEQ_1, FINSEQ_2, MATRIX_0, GROUP_4, MATRIX_3,
VECTSP_4, RELAT_1, LAPLACE, MATRIX13, FVSUM_1, ALGSTR_0, CARD_1,
ORDINAL1;
expansions TARSKI, XBOOLE_0, FINSEQ_1, MATRIX_1;
theorems CARD_1, CARD_2, FINSEQ_1, FINSEQ_2, FINSEQ_3, FINSEQOP, FINSOP_1,
FRECHET, FUNCT_1, FUNCT_2, FUNCT_7, FUNCOP_1, FVSUM_1, GROUP_1, LAPLACE,
MATRIX_0, MATRIX_3, MATRIX_4, MATRIX_5, MATRIX_6, MATRIX_7, MATRIX11,
MATRIX13, MATRIXR1, MATRIXR2, MATRLIN, NAT_1, ORDINAL1, PARTFUN1,
RELAT_1, RLVECT_1, STIRL2_1, STRUCT_0, TARSKI, WELLORD2, VECTSP_1,
VECTSP_4, VECTSP_6, VECTSP_7, VECTSP_9, XBOOLE_0, XBOOLE_1, XREAL_1,
XXREAL_0, ZFMISC_1, NAT_D, PRE_POLY, MEASURE6, MATRIX_1, VALUED_1;
schemes FINSEQ_2, NAT_1, MATRIX_0;
begin :: Preliminaries
reserve x,y for set,
i,j,k,l,m,n for Nat,
K for Field,
N for without_zero finite Subset of NAT,
a,b for Element of K,
A,B,B1,B2,X,X1,X2 for (Matrix of K),
A9 for (Matrix of m,n,K),
B9 for (Matrix of m,k,K);
theorem Th1:
width A = len B implies (a*A) * B = a * (A*B)
proof
set aA=a*A;
set AB=A*B;
set aAB=a*AB;
assume
A1: width A=len B;
then
A2: width AB=width B by MATRIX_3:def 4;
A3: len aAB=len AB by MATRIX_3:def 5;
then
A4: len aAB = len A by A1,MATRIX_3:def 4;
A5: width aA=width A by MATRIX_3:def 5;
then
A6: len aA=len A & len aA=len (aA*B) by A1,MATRIX_3:def 4,def 5;
A7: now
let i,j;
assume [i,j] in Indices aAB;
then
A8: [i,j] in Indices AB by MATRIXR1:18;
then i in dom AB by ZFMISC_1:87;
then i in Seg len A by A3,A4,FINSEQ_1:def 3;
then
A9: 1<=i & i<=len A by FINSEQ_1:1;
dom AB = Seg len AB by FINSEQ_1:def 3
.= dom (aA*B) by A3,A4,A6,FINSEQ_1:def 3;
then
A10: [i,j] in Indices (aA*B) by A1,A5,A2,A8,MATRIX_3:def 4;
thus aAB*(i,j) = a*(AB*(i,j)) by A8,MATRIX_3:def 5
.= a*(Line(A,i)"*"Col(B,j)) by A1,A8,MATRIX_3:def 4
.= Sum(a*mlt(Line(A,i),Col(B,j))) by FVSUM_1:73
.= Sum(mlt(a*Line(A,i),Col(B,j))) by A1,FVSUM_1:68
.= Line(aA,i)"*"Col(B,j) by A9,MATRIXR1:20
.= (aA*B)*(i,j) by A1,A5,A10,MATRIX_3:def 4;
end;
width aAB=width AB & width B=width (aA*B) by A1,A5,MATRIX_3:def 4,def 5;
hence thesis by A2,A4,A6,A7,MATRIX_0:21;
end;
theorem Th2:
1_K * A = A & a * (b * A) = (a * b) * A
proof
set 1A=1_K*A;
set bA=b*A;
set ab=a*b;
set abA=ab*A;
A1: now
let i,j such that
A2: [i,j] in Indices A;
thus A*(i,j) = 1_K*(A*(i,j)) by VECTSP_1:def 6
.= 1A*(i,j) by A2,MATRIX_3:def 5;
end;
A3: len(a*bA) = len bA by MATRIX_3:def 5
.= len A by MATRIX_3:def 5
.= len abA by MATRIX_3:def 5;
A4: now
A5: Indices bA=Indices A by MATRIXR1:18;
A6: Indices (a*bA)=Indices bA by MATRIXR1:18;
let i,j such that
A7: [i,j] in Indices (a*bA);
thus (a*bA)*(i,j) = a*(bA*(i,j)) by A7,A6,MATRIX_3:def 5
.= a*(b*(A*(i,j))) by A7,A6,A5,MATRIX_3:def 5
.= (a*b)*(A*(i,j)) by GROUP_1:def 3
.= abA*(i,j) by A7,A6,A5,MATRIX_3:def 5;
end;
A8: width(a*bA) = width bA by MATRIX_3:def 5
.= width A by MATRIX_3:def 5
.= width abA by MATRIX_3:def 5;
len 1A=len A & width 1A=width A by MATRIX_3:def 5;
hence thesis by A1,A3,A8,A4,MATRIX_0:21;
end;
Lm1: Indices A=Indices -A
proof
dom A = Seg len A by FINSEQ_1:def 3
.= Seg len (-A) by MATRIX_3:def 2
.= dom (-A) by FINSEQ_1:def 3;
hence thesis by MATRIX_3:def 2;
end;
Lm2: a <> 0.K implies power(K).(a,n)<>0.K
proof
defpred P[Nat] means for n st n=$1 holds power(K).(a,n)<>0.K;
assume
A1: a<>0.K;
A2: for k st P[k] holds P[k+1]
proof
let k;
assume P[k];
then
A3: power(K).(a,k) <> 0.K;
let n;
assume n=k+1;
then power(K).(a,n)=power(K).(a,k) * a by GROUP_1:def 7;
hence thesis by A1,A3,VECTSP_1:12;
end;
A4: P[0]
proof
A5: 1_K<>0.K;
let n;
assume n=0;
hence thesis by A5,GROUP_1:def 7;
end;
for k holds P[k] from NAT_1:sch 2(A4,A2);
hence thesis;
end;
theorem Th3:
for K be non empty addLoopStr, f,g,h,w be FinSequence of K st len
f=len g & len h = len w holds (f^h) + (g^w)= (f+g) ^ (h+w)
proof
let K be non empty addLoopStr, f,g,h,w be FinSequence of K such that
A1: len f=len g and
A2: len h = len w;
set KK=the carrier of K;
reconsider H=h,W=w as Element of (len h)-tuples_on KK by A2,FINSEQ_2:92;
reconsider F=f,G=g as Element of (len f)-tuples_on KK by A1,FINSEQ_2:92;
reconsider FH=F^H,GW=G^W,Th36W=(F+G)^(H+W) as Tuple of len f+len h, KK;
reconsider FH,GW,Th36W as Element of (len f+len h) -tuples_on KK
by FINSEQ_2:131;
now
let i such that
A3: i in Seg (len f+len h);
A4: i in dom FH by A3,FINSEQ_2:124;
now
per cases by A4,FINSEQ_1:25;
suppose
A5: i in dom f;
A6: rng f c= KK & rng g c= KK by RELAT_1:def 19;
A7: dom (F+G)=Seg len f by FINSEQ_2:124;
A8: f.i in rng f by A5,FUNCT_1:def 3;
A9: dom F=Seg len f by FINSEQ_2:124;
A10: dom G=Seg len f by FINSEQ_2:124;
then g.i in rng g by A5,A9,FUNCT_1:def 3;
then reconsider fi=f.i,gi=g.i as Element of K by A8,A6;
A11: FH.i=fi by A5,FINSEQ_1:def 7;
GW.i=gi by A5,A9,A10,FINSEQ_1:def 7;
hence (FH+GW).i = fi+gi by A3,A11,FVSUM_1:18
.= (F+G).i by A5,A9,FVSUM_1:18
.= Th36W.i by A5,A9,A7,FINSEQ_1:def 7;
end;
suppose
ex n st n in dom h & i=len f + n;
then consider n such that
A12: n in dom h and
A13: i=len f + n;
A14: h.n in rng h by A12,FUNCT_1:def 3;
A15: rng h c= KK & rng w c= KK by RELAT_1:def 19;
A16: dom H=Seg len h by FINSEQ_2:124;
A17: dom W=Seg len h by FINSEQ_2:124;
then w.n in rng w by A12,A16,FUNCT_1:def 3;
then reconsider hn=h.n,wn=w.n as Element of K by A14,A15;
A18: FH.i=hn by A12,A13,FINSEQ_1:def 7;
A19: dom (H+W)=Seg len h & len (F+G)=len f by CARD_1:def 7,FINSEQ_2:124;
GW.i=wn by A1,A12,A13,A16,A17,FINSEQ_1:def 7;
hence (FH+GW).i = hn+wn by A3,A18,FVSUM_1:18
.= (H+W).n by A12,A16,FVSUM_1:18
.= Th36W.i by A12,A13,A16,A19,FINSEQ_1:def 7;
end;
end;
hence (FH+GW).i=Th36W.i;
end;
hence thesis by FINSEQ_2:119;
end;
theorem Th4:
for K be non empty multMagma, f,g be FinSequence of K,a be
Element of K holds a * (f^g) = (a*f) ^ (a*g)
proof
let K be non empty multMagma, f,g be FinSequence of K,a be Element of K;
set KK=the carrier of K;
reconsider F=f as Element of (len f)-tuples_on KK by FINSEQ_2:92;
reconsider G=g as Element of (len g)-tuples_on KK by FINSEQ_2:92;
reconsider FG=F^G,aFaG=(a*F)^(a*G) as Element of (len f+len g)-tuples_on KK
by FINSEQ_2:131;
now
let i such that
A1: i in Seg (len f+len g);
A2: i in dom FG by A1,FINSEQ_2:124;
now
per cases by A2,FINSEQ_1:25;
suppose
A3: i in dom f;
A4: rng f c= KK by RELAT_1:def 19;
f.i in rng f by A3,FUNCT_1:def 3;
then reconsider fi=f.i as Element of K by A4;
A5: dom F=Seg len f by FINSEQ_2:124;
A6: dom (a*F)=Seg len f by FINSEQ_2:124;
FG.i=fi by A3,FINSEQ_1:def 7;
hence (a*FG).i = a*fi by A1,FVSUM_1:51
.= (a*F).i by A3,A5,FVSUM_1:51
.= aFaG.i by A3,A5,A6,FINSEQ_1:def 7;
end;
suppose
A7: ex n st n in dom g & i=len f + n;
A8: rng g c= KK by RELAT_1:def 19;
A9: dom (a*G)=Seg len g & len (a*F)=len f by CARD_1:def 7,FINSEQ_2:124;
consider n such that
A10: n in dom g and
A11: i=len f + n by A7;
g.n in rng g by A10,FUNCT_1:def 3;
then reconsider gn=g.n as Element of K by A8;
A12: dom G=Seg len g by FINSEQ_2:124;
FG.i=gn by A10,A11,FINSEQ_1:def 7;
hence (a*FG).i = a*gn by A1,FVSUM_1:51
.= (a*G).n by A10,A12,FVSUM_1:51
.= aFaG.i by A10,A11,A12,A9,FINSEQ_1:def 7;
end;
end;
hence (a*FG).i=aFaG.i;
end;
hence thesis by FINSEQ_2:119;
end;
theorem Th5:
for f be Function for p1,p2,f1,f2 be FinSequence st rng p1 c= dom
f & rng p2 c= dom f & f1=f*p1 & f2=f*p2 holds f * (p1 ^ p2) = f1 ^ f2
proof
let f be Function;
let p1,p2,f1,f2 be FinSequence such that
A1: rng p1 c= dom f and
A2: rng p2 c= dom f and
A3: f1=f*p1 and
A4: f2=f*p2;
A5: dom (p1^p2)=Seg (len p1+len p2) by FINSEQ_1:def 7;
rng (p1^p2) = rng p1 \/rng p2 by FINSEQ_1:31;
then
A6: dom (f*(p1^p2))=dom (p1^p2) by A1,A2,RELAT_1:27,XBOOLE_1:8;
A7: dom f1=dom p1 by A1,A3,RELAT_1:27;
then
A8: len f1=len p1 by FINSEQ_3:29;
A9: dom f2=dom p2 by A2,A4,RELAT_1:27;
then len f2=len p2 by FINSEQ_3:29;
then
A10: dom (f1^f2)=dom (f*(p1^p2)) by A6,A8,A5,FINSEQ_1:def 7;
now
let x be object such that
A11: x in dom (f1^f2);
reconsider i=x as Element of NAT by A11;
now
per cases;
suppose
A12: i in dom p1;
hence (f1^f2).i = f1.i by A7,FINSEQ_1:def 7
.= f.(p1.i) by A3,A7,A12,FUNCT_1:12
.= f.((p1^p2).i) by A12,FINSEQ_1:def 7
.= (f*(p1^p2)).i by A10,A11,FUNCT_1:12;
end;
suppose
not i in dom p1;
then consider n such that
A13: n in dom p2 and
A14: i=len p1 + n by A7,A9,A8,A11,FINSEQ_1:25;
thus (f1^f2).i = f2.n by A9,A8,A13,A14,FINSEQ_1:def 7
.= f.(p2.n) by A4,A9,A13,FUNCT_1:12
.= f.((p1^p2).i) by A13,A14,FINSEQ_1:def 7
.= (f*(p1^p2)).i by A10,A11,FUNCT_1:12;
end;
end;
hence (f1^f2).x = (f*(p1^p2)).x;
end;
hence thesis by A10,FUNCT_1:2;
end;
theorem Th6:
for f be FinSequence of NAT,n st f is one-to-one & rng f c= Seg n
& for i,j st i in dom f & j in dom f & i < j holds f.i < f.j holds Sgm rng f =
f
proof
defpred P[Nat] means for f be FinSequence of NAT,n st len f=$1 & rng f c=
Seg n & f is one-to-one & for i,j st i in dom f & j in dom f & i < j holds f.i
< f.j holds Sgm rng f = f;
A1: P[0]
proof
let f be FinSequence of NAT,n such that
A2: len f=0 and
rng f c= Seg n and
f is one-to-one and
for i,j st i in dom f & j in dom f & i < j holds f.i < f.j;
f={} by A2;
hence thesis by FINSEQ_3:43;
end;
A3: for n st P[n] holds P[n+1]
proof
let n such that
A4: P[n];
set n1=n+1;
let f be FinSequence of NAT,k such that
A5: len f=n1 and
A6: rng f c= Seg k and
A7: f is one-to-one and
A8: for i,j st i in dom f & j in dom f & i < j holds f.i < f.j;
set fn=f|n;
A9: f=fn^<*f.n1*> by A5,FINSEQ_3:55;
then
A10: rng fn c= rng f by FINSEQ_1:29;
A11: dom fn c= dom f by A9,FINSEQ_1:26;
A12: for i,j st i in dom fn & j in dom fn & i < j holds fn.i < fn.j
proof
let i,j such that
A13: i in dom fn & j in dom fn and
A14: i < j;
fn.i=f.i & fn.j=f.j by A9,A13,FINSEQ_1:def 7;
hence thesis by A8,A11,A13,A14;
end;
A15: len fn =n by A5,FINSEQ_3:53;
A16: now
A17: n1 in Seg n1 & dom f=Seg n1 by A5,FINSEQ_1:4,def 3;
let m9,n9 being Nat such that
A18: m9 in rng fn and
A19: n9 in {f.n1};
consider x being object such that
A20: x in dom fn and
A21: fn.x=m9 by A18,FUNCT_1:def 3;
reconsider x as Element of NAT by A20;
A22: f.x = fn.x by A9,A20,FINSEQ_1:def 7;
dom fn=Seg n by A15,FINSEQ_1:def 3;
then x<=n by A20,FINSEQ_1:1;
then x < n1 by NAT_1:13;
then f.x < f.n1 by A8,A11,A20,A17;
hence m9 < n9 by A19,A21,A22,TARSKI:def 1;
end;
fn is one-to-one by A7,FUNCT_1:52;
then
A23: Sgm rng fn = fn by A4,A6,A15,A10,A12,XBOOLE_1:1;
A24: rng <*f.n1*>={f.n1} by FINSEQ_1:39;
rng <*f.n1*> c= rng f by A9,FINSEQ_1:30;
then
A25: {f.n1} c= Seg k by A6,A24;
A26: rng f = rng fn\/rng<*f.n1*> by A9,FINSEQ_1:31;
A27: f.n1 in {f.n1} by TARSKI:def 1;
rng fn c= Seg k by A6,A10;
hence Sgm rng f = fn ^ Sgm {f.n1} by A26,A24,A25,A23,A16,FINSEQ_3:42
.= f by A9,A25,A27,FINSEQ_3:44;
end;
let f be FinSequence of NAT,n such that
A28: f is one-to-one & rng f c= Seg n & for i,j st i in dom f & j in dom
f & i < j holds f.i < f.j;
for n holds P[n] from NAT_1:sch 2(A1,A3);
then for g be FinSequence of NAT,n st len g=len f & rng g c= Seg n & g is
one-to-one & for i,j st i in dom g & j in dom g & i < j holds g.i < g.j holds
Sgm rng g = g;
hence thesis by A28;
end;
theorem Th7:
for K be Abelian add-associative right_zeroed
right_complementable non empty addLoopStr for p be FinSequence of K for i,j
st i in dom p & j in dom p & i<>j & for k st k in dom p & k<>i & k<>j holds p.k
=0.K holds Sum p = p/.i+p/.j
proof
let K be Abelian add-associative right_zeroed right_complementable non
empty addLoopStr;
let p be FinSequence of K;
A1: now
let i,j such that
A2: i in dom p and
A3: j in dom p and
A4: ii & k<>j holds p.k=0.K;
A6: dom p=Seg len p by FINSEQ_1:def 3;
then i in NAT & 1 <= i by A2,FINSEQ_1:1;
then
A7: i in Seg i;
set pI=p|i;
consider q be FinSequence such that
A8: p = pI^q by FINSEQ_1:80;
reconsider q as FinSequence of K by A8,FINSEQ_1:36;
A9: i <= len p by A2,A6,FINSEQ_1:1;
then
A10: len pI=i by FINSEQ_1:17;
A11: dom pI=Seg i by A9,FINSEQ_1:17;
then not j in dom pI by A4,FINSEQ_1:1;
then consider ji be Nat such that
A12: ji in dom q and
A13: j=i + ji by A3,A8,A10,FINSEQ_1:25;
now
let k such that
A14: k in dom q and
A15: k<>ji;
reconsider kk=k as Element of NAT by ORDINAL1:def 12;
A16: i+kk <> i+ji by A15;
dom q = Seg len q by FINSEQ_1:def 3;
then k >= 1 by A14,FINSEQ_1:1;
then k+i >= i+1 by XREAL_1:7;
then
A17: i+kk<>i by NAT_1:13;
thus q.k = p.(i+kk) by A8,A10,A14,FINSEQ_1:def 7
.= 0.K by A5,A8,A10,A13,A14,A17,A16,FINSEQ_1:28;
end;
then
A18: Sum q = q.ji by A12,MATRIX_3:12
.= p.j by A8,A10,A12,A13,FINSEQ_1:def 7
.= p/.j by A3,PARTFUN1:def 6;
A19: Seg i c= Seg len p by A9,FINSEQ_1:5;
now
let k such that
A20: k in dom pI and
A21: k<>i;
reconsider kk=k as Element of NAT by ORDINAL1:def 12;
A22: k<>j by A4,A11,A20,FINSEQ_1:1;
thus pI.k = p.kk by A8,A20,FINSEQ_1:def 7
.= 0.K by A5,A6,A11,A19,A20,A21,A22;
end;
then Sum pI = pI.i by A7,A11,MATRIX_3:12
.= p.i by A8,A7,A11,FINSEQ_1:def 7
.= p/.i by A6,A7,A19,PARTFUN1:def 6;
hence Sum p=p/.i+p/.j by A8,A18,RLVECT_1:41;
end;
let i,j such that
A23: i in dom p & j in dom p and
A24: i<>j and
A25: for k st k in dom p & k<>i & k<>j holds p.k=0.K;
A26: ij & k<>i holds p.k=0.K by A25;
hence thesis by A1,A23,A25,A26;
end;
theorem Th8:
i in Seg m implies Sgm (Seg (n+m)\Seg n).i=n+i
proof
assume
A1: i in Seg m;
reconsider N=n as Element of NAT by ORDINAL1:def 12;
set I=idseq m;
A2: dom I=Seg len I by FINSEQ_1:def 3;
A3: dom (N Shift I) = {k+N where k is Nat: k in dom I} by VALUED_1:def 12;
A4: Seg (n+m)\Seg n c= dom (N Shift I)
proof
let x be object such that
A5: x in Seg (n+m)\Seg n;
reconsider i=x as Element of NAT by A5;
A6: i in Seg (n+m) by A5,XBOOLE_0:def 5;
not i in Seg n by A5,XBOOLE_0:def 5;
then
A7: i<1 or i>n;
then reconsider IN=i-n as Element of NAT by A6,FINSEQ_1:1,NAT_1:21;
A8: n+IN=i;
i<=n+m by A6,FINSEQ_1:1;
then
A9: IN<=m by A8,XREAL_1:8;
IN>=1 by A6,A7,A8,FINSEQ_1:1,NAT_1:19;
then IN in dom I by A9;
then n+IN in dom (N Shift I) by A3;
hence thesis;
end;
dom (N Shift I) c=Seg (n+m)\Seg n
proof
let x be object;
assume x in dom (N Shift I);
then consider k be Nat such that
A10: k+n=x and
A11: k in dom I by A3;
k<=m by A11,FINSEQ_1:1;
then
A12: n+k<=n+m by XREAL_1:7;
1<=k by A11,FINSEQ_1:1;
then
A13: n+1<=n+k by XREAL_1:7;
then n+k >n by NAT_1:13;
then
A14: not k+n in Seg n by FINSEQ_1:1;
1<=n+1 by NAT_1:11;
then 1<=n+k by A13,XXREAL_0:2;
then n+k in Seg (n+m) by A12;
hence thesis by A10,A14,XBOOLE_0:def 5;
end;
then Seg (n+m)\Seg n = dom (N Shift I) by A4;
hence thesis by A1,A2,VALUED_1:44;
end;
theorem Th9:
for D be non empty set, A be Matrix of D for Bx,By,Cx,Cy be
without_zero finite Subset of NAT st [:Bx,By:] c= Indices A & [:Cx,Cy:] c=
Indices A for B be Matrix of card Bx,card By,D, C be Matrix of card Cx,card Cy,
D st for i,j,bi,bj,ci,cj be Nat st [i,j] in [:Bx,By:]/\[:Cx,Cy:] & bi = Sgm Bx"
.i & bj = Sgm By".j & ci = Sgm Cx".i & cj = Sgm Cy".j holds B*(bi,bj) = C*(ci,
cj) ex M be Matrix of len A,width A,D st Segm(M,Bx,By) = B & Segm(M,Cx,Cy) = C
& for i,j st [i,j] in Indices M\([:Bx,By:]\/[:Cx,Cy:]) holds M*(i,j) = A*(i,j)
proof
let D be non empty set, A be Matrix of D;
let Bx,By,Cx,Cy be without_zero finite Subset of NAT such that
A1: [:Bx,By:] c= Indices A and
A2: [:Cx,Cy:] c= Indices A;
set w=width A;
set l=len A;
set cy=card Cy;
set cx=card Cx;
set bY=card By;
set bx=card Bx;
let B be Matrix of bx,bY,D,C be Matrix of cx,cy,D such that
A3: for i,j,bi,bj,ci,cj be Nat st [i,j] in [:Bx,By:]/\[:Cx,Cy:] & bi =
Sgm Bx".i & bj = Sgm By".j & ci = Sgm Cx".i & cj = Sgm Cy".j holds B*(bi,bj) =
C*(ci,cj);
A4: ex kBy be Nat st By c= Seg kBy by MATRIX13:43;
then
A5: rng Sgm By=By by FINSEQ_1:def 13;
defpred P[set,set,set] means for i,j st $1=i & $2=j holds ([i,j] in [:Bx,By
:] implies (ex m,n st m in dom Sgm Bx & n in dom Sgm By & Sgm Bx.m=i & Sgm By.n
=j)& for m,n st m in dom Sgm Bx & n in dom Sgm By & Sgm Bx.m=i & Sgm By.n=j
holds $3=B*(m,n))& ([i,j] in [:Cx,Cy:] implies (ex m,n st m in dom Sgm Cx & n
in dom Sgm Cy & Sgm Cx.m=i & Sgm Cy.n=j)&for m,n st m in dom Sgm Cx & n in dom
Sgm Cy & Sgm Cx.m=i & Sgm Cy.n=j holds $3=C*(m,n))& (not [i,j] in [:Bx,By:]\/[:
Cx,Cy:] implies $3=A*(i,j));
A6: ex kBx be Nat st Bx c= Seg kBx by MATRIX13:43;
then
A7: rng Sgm Bx=Bx by FINSEQ_1:def 13;
A8: dom Sgm By=Seg bY by A4,FINSEQ_3:40;
A9: Sgm By is one-to-one by A4,FINSEQ_3:92;
A10: Sgm Bx is one-to-one by A6,FINSEQ_3:92;
A11: ex kCy be Nat st Cy c= Seg kCy by MATRIX13:43;
then
A12: rng Sgm Cy=Cy by FINSEQ_1:def 13;
A13: Sgm Cy is one-to-one by A11,FINSEQ_3:92;
A14: ex kCx be Nat st Cx c= Seg kCx by MATRIX13:43;
then
A15: rng Sgm Cx=Cx by FINSEQ_1:def 13;
A16: Sgm Cx is one-to-one by A14,FINSEQ_3:92;
A17: for i,j st [i,j] in [:Seg l, Seg w:] ex x being Element of D st P[i,j,x ]
proof
let i,j such that
[i,j] in [:Seg l, Seg w:];
per cases;
suppose
A18: [i,j] in [:Bx,By:] & [i,j] in [:Cx,Cy:];
then j in Cy by ZFMISC_1:87;
then consider yC be object such that
A19: yC in dom Sgm Cy and
A20: Sgm Cy.yC=j by A12,FUNCT_1:def 3;
j in By by A18,ZFMISC_1:87;
then consider y being object such that
A21: y in dom Sgm By and
A22: Sgm By.y=j by A5,FUNCT_1:def 3;
i in Cx by A18,ZFMISC_1:87;
then consider xC be object such that
A23: xC in dom Sgm Cx and
A24: Sgm Cx.xC=i by A15,FUNCT_1:def 3;
i in Bx by A18,ZFMISC_1:87;
then consider x being object such that
A25: x in dom Sgm Bx and
A26: Sgm Bx.x=i by A7,FUNCT_1:def 3;
reconsider x,y as Element of NAT by A25,A21;
take BB=B*(x,y);
A27: now
let m,n;
assume m in dom Sgm Cx & n in dom Sgm Cy & Sgm Cx.m=i & Sgm Cy.n=j;
then
A28: Sgm Cx".i=m & Sgm Cy".j=n by A16,A13,FUNCT_1:32;
A29: Sgm By".j=y by A9,A21,A22,FUNCT_1:32;
[i,j] in [:Bx,By:]/\[:Cx,Cy:] & Sgm Bx".i=x by A10,A18,A25,A26,
FUNCT_1:32,XBOOLE_0:def 4;
hence BB=C*(m,n) by A3,A29,A28;
end;
now
let m,n such that
A30: m in dom Sgm Bx and
A31: n in dom Sgm By and
A32: Sgm Bx.m=i and
A33: Sgm By.n=j;
x=m by A10,A25,A26,A30,A32,FUNCT_1:def 4;
hence BB=B*(m,n) by A9,A21,A22,A31,A33,FUNCT_1:def 4;
end;
hence thesis by A18,A25,A26,A21,A22,A23,A24,A19,A20,A27,XBOOLE_0:def 3;
end;
suppose
A34: [i,j] in [:Bx,By:] & not [i,j] in [:Cx,Cy:];
then j in By by ZFMISC_1:87;
then consider y being object such that
A35: y in dom Sgm By and
A36: Sgm By.y=j by A5,FUNCT_1:def 3;
i in Bx by A34,ZFMISC_1:87;
then consider x being object such that
A37: x in dom Sgm Bx and
A38: Sgm Bx.x=i by A7,FUNCT_1:def 3;
reconsider x,y as Element of NAT by A37,A35;
take BB=B*(x,y);
now
let m,n such that
A39: m in dom Sgm Bx and
A40: n in dom Sgm By and
A41: Sgm Bx.m=i and
A42: Sgm By.n=j;
x = m by A10,A37,A38,A39,A41,FUNCT_1:def 4;
hence BB=B*(m,n) by A9,A35,A36,A40,A42,FUNCT_1:def 4;
end;
hence thesis by A34,A37,A38,A35,A36,XBOOLE_0:def 3;
end;
suppose
A43: not [i,j] in [:Bx,By:] & [i,j] in [:Cx,Cy:];
then j in Cy by ZFMISC_1:87;
then consider y being object such that
A44: y in dom Sgm Cy and
A45: Sgm Cy.y=j by A12,FUNCT_1:def 3;
i in Cx by A43,ZFMISC_1:87;
then consider x being object such that
A46: x in dom Sgm Cx and
A47: Sgm Cx.x=i by A15,FUNCT_1:def 3;
reconsider x,y as Element of NAT by A46,A44;
take CC=C*(x,y);
now
let m,n such that
A48: m in dom Sgm Cx and
A49: n in dom Sgm Cy and
A50: Sgm Cx.m=i and
A51: Sgm Cy.n=j;
x = m by A16,A46,A47,A48,A50,FUNCT_1:def 4;
hence CC=C*(m,n) by A13,A44,A45,A49,A51,FUNCT_1:def 4;
end;
hence thesis by A43,A46,A47,A44,A45,XBOOLE_0:def 3;
end;
suppose
A52: not [i,j] in [:Bx,By:] & not [i,j] in [:Cx,Cy:];
take A*(i,j);
thus thesis by A52;
end;
end;
consider M be Matrix of l,w,D such that
A53: for i,j st [i,j] in Indices M holds P[i,j,M*(i,j)] from MATRIX_0:
sch 2(A17);
set MB=Segm(M,Bx,By);
take M;
A is Matrix of l,w,D by MATRIX_0:51;
then
A54: Indices A=Indices M by MATRIX_0:26;
A55: dom Sgm Bx=Seg bx by A6,FINSEQ_3:40;
now
let i,j such that
A56: [i,j] in Indices MB;
bx<>0 by A56,MATRIX_0:22;
then
A57: Indices MB=[:Seg bx,Seg bY:] by MATRIX_0:23;
then
A58: j in Seg bY by A56,ZFMISC_1:87;
then
A59: Sgm By.j in By by A5,A8,FUNCT_1:def 3;
A60: i in Seg bx by A56,A57,ZFMISC_1:87;
then Sgm Bx.i in Bx by A7,A55,FUNCT_1:def 3;
then [Sgm Bx.i,Sgm By.j] in [:Bx,By:] by A59,ZFMISC_1:87;
hence B*(i,j) = M*(Sgm Bx.i,Sgm By.j) by A1,A55,A8,A53,A54,A60,A58
.= MB*(i,j) by A56,MATRIX13:def 1;
end;
hence MB=B by MATRIX_0:27;
set MC=Segm(M,Cx,Cy);
A61: dom Sgm Cy=Seg cy by A11,FINSEQ_3:40;
A62: dom Sgm Cx=Seg cx by A14,FINSEQ_3:40;
now
let i,j such that
A63: [i,j] in Indices MC;
cx<>0 by A63,MATRIX_0:22;
then
A64: Indices MC=[:Seg cx,Seg cy:] by MATRIX_0:23;
then
A65: j in Seg cy by A63,ZFMISC_1:87;
then
A66: Sgm Cy.j in Cy by A12,A61,FUNCT_1:def 3;
A67: i in Seg cx by A63,A64,ZFMISC_1:87;
then Sgm Cx.i in Cx by A15,A62,FUNCT_1:def 3;
then [Sgm Cx.i,Sgm Cy.j] in [:Cx,Cy:] by A66,ZFMISC_1:87;
hence C*(i,j) = M*(Sgm Cx.i,Sgm Cy.j) by A2,A62,A61,A53,A54,A67,A65
.= MC*(i,j) by A63,MATRIX13:def 1;
end;
hence MC=C by MATRIX_0:27;
let i,j such that
A68: [i,j] in Indices M\([:Bx,By:]\/[:Cx,Cy:]);
not [i,j] in [:Bx,By:]\/[:Cx,Cy:] by A68,XBOOLE_0:def 5;
hence thesis by A53,A68;
end;
theorem Th10:
for P,Q,Q9 be without_zero finite Subset of NAT st [:P,Q9:] c=
Indices A for i,j st i in dom A \ P & j in Seg width A \ Q & A*(i,j) <> 0.K & Q
c= Q9 & Line(A,i) * Sgm Q9 = card Q9 |-> 0.K holds the_rank_of A > the_rank_of
Segm(A,P,Q)
proof
let P,Q,R be without_zero finite Subset of NAT such that
A1: [:P,R:] c= Indices A;
let i,j such that
A2: i in dom A \ P and
A3: j in Seg width A \ Q and
A4: A*(i,j)<> 0.K and
A5: Q c= R and
A6: Line(A,i) * Sgm R = card R |-> 0.K;
A7: dom A=Seg len A by FINSEQ_1:def 3;
then
A8: i in Seg len A by A2,XBOOLE_0:def 5;
A9: [:P,Q:] c= [:P,R:] by A5,ZFMISC_1:95;
then
A10: [:P,Q:] c= Indices A by A1;
reconsider i0=i,j0=j as non zero Element of NAT by A2,A3,A7;
A11: j in Seg width A by A3,XBOOLE_0:def 5;
set S=Segm(A,P,Q);
consider P9,Q9 be without_zero finite Subset of NAT such that
A12: [:P9,Q9:] c= Indices S and
A13: card P9 = card Q9 and
A14: card P9 = the_rank_of S and
A15: Det EqSegm(S,P9,Q9)<>0.K by MATRIX13:def 4;
P9={} iff Q9={} by A13;
then consider P2,Q2 be without_zero finite Subset of NAT such that
A16: P2 c= P and
A17: Q2 c= Q and
P2 = Sgm P.:P9 and
Q2=Sgm Q.:Q9 and
A18: card P2=card P9 and
A19: card Q2=card Q9 and
A20: Segm(S,P9,Q9) = Segm(A,P2,Q2) by A12,MATRIX13:57;
set Q2j=Q2\/{j0};
set P2i=P2\/{i0};
set ESS=EqSegm(A,P2i,Q2j);
set SS=Segm(A,P2i,Q2j);
per cases;
suppose
[:P,Q:] ={};
then card P=0 or card Q=0 by CARD_1:27,ZFMISC_1:90;
then
A21: the_rank_of S=0 by MATRIX13:77;
[i,j] in Indices A by A7,A8,A11,ZFMISC_1:87;
hence thesis by A4,A21,MATRIX13:94;
end;
suppose
A22: [:P,Q:]<>{};
then P c= dom A by A10,ZFMISC_1:114;
then
A23: P2 c= dom A by A16;
[:P,R:]<>{} by A9,A22,XBOOLE_1:3;
then
A24: R c= Seg width A by A1,ZFMISC_1:114;
then
A25: dom Sgm R=Seg card R by FINSEQ_3:40;
Q c= Seg width A by A10,A22,ZFMISC_1:114;
then
A26: Q2 c= Seg width A by A17;
A27: {j0}c= Seg width A by A11,ZFMISC_1:31;
then
A28: Sgm Q2j is one-to-one by A26,FINSEQ_3:92,XBOOLE_1:8;
A29: Q2j c= Seg width A by A26,A27,XBOOLE_1:8;
then
A30: rng Sgm Q2j=Q2j by FINSEQ_1:def 13;
A31: {i0} c= dom A by A7,A8,ZFMISC_1:31;
then
A32: P2i c= dom A by A23,XBOOLE_1:8;
then
A33: [:P2i,Q2j:] c= Indices A by A29,ZFMISC_1:96;
A34: dom Sgm P2i=Seg card P2i by A7,A23,A31,FINSEQ_3:40,XBOOLE_1:8;
i in {i} by TARSKI:def 1;
then
A35: i in P2i by XBOOLE_0:def 3;
A36: not i in P2 by A2,A16,XBOOLE_0:def 5;
then
A37: card P2i=card P2+1 by CARD_2:41;
then
A38: card P2i-'1=card P9 by A18,NAT_D:34;
A39: not j in Q2 by A3,A17,XBOOLE_0:def 5;
then
A40: card Q2j=card Q2+1 by CARD_2:41;
then
A41: ESS = SS by A13,A18,A19,A36,CARD_2:41,MATRIX13:def 3;
j in {j} by TARSKI:def 1;
then j in Q2j by XBOOLE_0:def 3;
then consider y being object such that
A42: y in dom Sgm Q2j and
A43: Sgm Q2j.y=j by A30,FUNCT_1:def 3;
rng Sgm P2i=P2i by A7,A32,FINSEQ_1:def 13;
then consider x being object such that
A44: x in dom Sgm P2i and
A45: Sgm P2i.x=i by A35,FUNCT_1:def 3;
reconsider x,y as Element of NAT by A44,A42;
-1_K<>0.K by VECTSP_1:28;
then
A46: power(K).(-1_K,x+y)<>0.K by Lm2;
set L=LaplaceExpL(ESS,x);
A47: dom L = Seg len L by FINSEQ_1:def 3
.= Seg card P2i by LAPLACE:def 7;
then
A48: y in dom L by A13,A18,A19,A26,A27,A37,A40,A42,FINSEQ_3:40,XBOOLE_1:8;
A49: dom Sgm Q2j=Seg card Q2j by A26,A27,FINSEQ_3:40,XBOOLE_1:8;
then Delete(ESS,x,y) = EqSegm(A,P2i\{i},Q2j\{j}) by A13,A18,A19,A37,A40,A34
,A44,A45,A42,A43,MATRIX13:64
.= EqSegm(A,P2,Q2j\{j}) by A36,ZFMISC_1:117
.= EqSegm(A,P2,Q2) by A39,ZFMISC_1:117
.= Segm(A,P2,Q2) by A13,A18,A19,MATRIX13:def 3
.= EqSegm(S,P9,Q9) by A13,A20,MATRIX13:def 3;
then
A50: power(K).(-1_K,x+y) * Det Delete(ESS,x,y)<>0.K by A15,A38,A46,VECTSP_1:12;
A51: Indices ESS = [:Seg card P2i,Seg card P2i:] by MATRIX_0:24;
then
A52: [x,y] in Indices ESS by A13,A18,A19,A37,A40,A34,A49,A44,A42,ZFMISC_1:87;
A53: rng Sgm R=R by A24,FINSEQ_1:def 13;
now
let k such that
A54: k in dom L and
A55: k <> y;
Sgm Q2j.k <> j by A13,A18,A19,A37,A40,A49,A28,A42,A43,A47,A54,A55,
FUNCT_1:def 4;
then
A56: not Sgm Q2j.k in {j} by TARSKI:def 1;
Sgm Q2j.k in Q2j by A13,A18,A19,A37,A40,A30,A49,A47,A54,FUNCT_1:def 3;
then Sgm Q2j.k in Q2 by A56,XBOOLE_0:def 3;
then
A57: Sgm Q2j.k in Q by A17;
then
A58: Sgm Q2j.k in R by A5;
consider z being object such that
A59: z in dom Sgm R and
A60: Sgm R.z=Sgm Q2j.k by A5,A53,A57,FUNCT_1:def 3;
reconsider z as Element of NAT by A59;
[x,k] in Indices ESS by A34,A44,A47,A51,A54,ZFMISC_1:87;
then ESS*(x,k) = A*(i,Sgm Q2j.k) by A45,A41,MATRIX13:def 1
.= Line(A,i).(Sgm R.z) by A24,A60,A58,MATRIX_0:def 7
.= (card R |-> 0.K).z by A6,A59,FUNCT_1:13
.= 0.K by A25,A59,FINSEQ_2:57;
hence L.k = 0.K*Cofactor(ESS,x,k) by A54,LAPLACE:def 7
.= 0.K;
end;
then
A61: L.y = Sum L by A48,MATRIX_3:12
.= Det ESS by A34,A44,LAPLACE:25;
L.y = SS*(x,y)*Cofactor(ESS,x,y) by A48,A41,LAPLACE:def 7
.= A*(i,j)*(power(K).(-1_K,x+y)*Det Delete(ESS,x,y)) by A45,A43,A41,A52,
MATRIX13:def 1;
then Det ESS<>0.K by A4,A61,A50,VECTSP_1:12;
then the_rank_of A >= card P2i by A13,A18,A19,A37,A40,A33,MATRIX13:def 4;
hence thesis by A14,A18,A37,NAT_1:13;
end;
end;
theorem Th11:
for N st N c= dom A & for i st i in dom A \ N holds Line(A,i) =
width A |-> 0.K holds the_rank_of A=the_rank_of Segm(A,N,Seg width A)
proof
let N such that
A1: N c= dom A and
A2: for i st i in (dom A) \ N holds Line(A,i) = width A |-> 0.K;
set w=width A;
set l=len A;
reconsider A9=A as Matrix of len A,width A,K by MATRIX_0:51;
set S=Segm(A9,N,Seg width A9);
consider U be finite Subset of w-VectSp_over K such that
A3: U is linearly-independent and
A4: U c= lines A9 and
A5: card U = the_rank_of A9 by MATRIX13:123;
A6: U c= lines S
proof
let x be object such that
A7: x in U;
consider Ni be Nat such that
A8: Ni in Seg l and
A9: x = Line(A9,Ni) by A4,A7,MATRIX13:103;
A10: dom A=Seg l by FINSEQ_1:def 3;
A11: Ni in N
proof
assume not Ni in N;
then Ni in dom A\N by A8,A10,XBOOLE_0:def 5;
then x = w |-> 0.K by A2,A9
.= 0.(w-VectSp_over K) by MATRIX13:102;
hence thesis by A3,A7,VECTSP_7:2;
end;
rng Sgm N=N by A1,A10,FINSEQ_1:def 13;
then consider i be object such that
A12: i in dom Sgm N and
A13: Sgm N.i=Ni by A11,FUNCT_1:def 3;
reconsider i as Element of NAT by A12;
A14: dom Sgm N=Seg card N by A1,A10,FINSEQ_3:40;
then Line(S,i) = x by A9,A12,A13,MATRIX13:48;
hence thesis by A12,A14,MATRIX13:103;
end;
A15: now
let W be finite Subset of w-VectSp_over K such that
A16: W is linearly-independent and
A17: W c= lines S;
dom A=Seg l by FINSEQ_1:def 3;
then lines S c= lines A9 by A1,MATRIX13:118;
then W c= lines A9 by A17;
hence card W <= the_rank_of A9 by A16,MATRIX13:123;
end;
w=card Seg w by FINSEQ_1:57;
hence thesis by A3,A5,A6,A15,MATRIX13:123;
end;
theorem Th12:
for N st N c= Seg width A & for i st i in Seg width A \ N holds
Col(A,i) = len A |-> 0.K holds the_rank_of A = the_rank_of Segm(A,Seg len A,N)
proof
let N such that
A1: N c= Seg width A and
A2: for i st i in Seg width A \ N holds Col(A,i) = len A |-> 0.K;
A3: dom A=Seg len A by FINSEQ_1:def 3;
per cases;
suppose
A4: card N=0;
the_rank_of A = 0
proof
A5: N={} by A4;
assume the_rank_of A <>0;
then consider i,j such that
A6: [i,j] in Indices A and
A7: A*(i,j) <> 0.K by MATRIX13:94;
A8: j in Seg width A by A6,ZFMISC_1:87;
A9: i in dom A by A6,ZFMISC_1:87;
then 0.K = (len A|-> 0.K).i by A3,FINSEQ_2:57
.= Col(A,j).i by A2,A8,A5
.= A*(i,j) by A9,MATRIX_0:def 8;
hence thesis by A7;
end;
hence thesis by A4,MATRIX13:77;
end;
suppose
A10: card N>0;
then N <> {};
then Seg width A <> {} by A1,XBOOLE_1:3;
then
A11: width A<>0;
then
A12: len A<>0 by MATRIX_0:def 3;
set AT=A@;
A13: [:dom A,N:] c= Indices A by A1,ZFMISC_1:95;
A14: width AT=len A by A11,MATRIX_0:54;
A15: len AT=width A by A11,MATRIX_0:54;
then
A16: N c= dom AT by A1,FINSEQ_1:def 3;
A17: dom AT=Seg len AT by FINSEQ_1:def 3;
A18: now
let i such that
A19: i in dom AT \ N;
i in dom AT by A19,XBOOLE_0:def 5;
hence Line(AT,i) = Col(A,i) by A15,A17,MATRIX_0:59
.= width AT |-> 0.K by A2,A15,A14,A17,A19;
end;
thus the_rank_of A = the_rank_of AT by MATRIX13:84
.= the_rank_of Segm(AT, N,Seg len A) by A14,A16,A18,Th11
.= the_rank_of (Segm(A,Seg len A,N)@) by A3,A10,A12,A13,MATRIX13:61
.= the_rank_of Segm(A,Seg len A,N) by MATRIX13:84;
end;
end;
theorem Th13:
for V be VectSp of K for U be finite Subset of V for u, v be
Vector of V,a st u in U & v in U holds Lin (U \ {u} \/ {u+a*v}) is Subspace of
Lin U
proof
let V be VectSp of K;
let U be finite Subset of V;
let u, v be Vector of V,a such that
A1: u in U & v in U;
set ua=u+a*v;
set UU=U\{u};
UU \/ {ua} c= the carrier of Lin U
proof
let x be object such that
A2: x in UU \/ {ua};
per cases by A2,XBOOLE_0:def 3;
suppose
x in UU;
then x in U by XBOOLE_0:def 5;
then x in Lin U by VECTSP_7:8;
hence thesis by STRUCT_0:def 5;
end;
suppose
A3: x in {ua};
A4: u in Lin U & a*v in Lin U by A1,VECTSP_4:21,VECTSP_7:8;
x=ua by A3,TARSKI:def 1;
then x in Lin U by A4,VECTSP_4:20;
hence thesis by STRUCT_0:def 5;
end;
end;
hence thesis by VECTSP_9:16;
end;
theorem Th14:
for V be VectSp of K for U be finite Subset of V for u, v be
Vector of V,a st u in U & v in U & (u = v implies (a <> -1_K or u = 0.V)) holds
Lin (U \ {u} \/ {u+a*v}) = Lin U
proof
let V be VectSp of K;
let U be finite Subset of V;
let u, v be Vector of V,a such that
A1: u in U and
A2: v in U and
A3: u = v implies (a <> -1_K or u = 0.V);
set ua=u+a*v;
set UU=U\{u};
U c= the carrier of Lin (UU\/{ua})
proof
let x be object such that
A4: x in U;
per cases;
suppose
A5: x=u;
per cases;
suppose
A6: u<>v;
A7: ua+(-a)*v = ua-a*v by VECTSP_1:21
.= u+(a*v-a*v) by RLVECT_1:def 3
.= u+0.V by VECTSP_1:16
.= u by RLVECT_1:def 4;
ua in {ua} by TARSKI:def 1;
then ua in UU\/{ua} by XBOOLE_0:def 3;
then
A8: ua in Lin(UU\/{ua}) by VECTSP_7:8;
v in UU by A2,A6,ZFMISC_1:56;
then v in UU\/{ua} by XBOOLE_0:def 3;
then (-a)*v in Lin(UU\/{ua}) by VECTSP_4:21,VECTSP_7:8;
then ua+(-a)*v in Lin(UU\/{ua}) by A8,VECTSP_4:20;
hence thesis by A5,A7,STRUCT_0:def 5;
end;
suppose
A9: u=v;
per cases by A3,A9;
suppose
a<>-1_K;
then 0.K<> -1_K-a by VECTSP_1:19;
then 0.K<> -(-1_K-a) by VECTSP_1:28;
then
A10: 0.K<>a+1_K by VECTSP_1:31;
ua in {ua} by TARSKI:def 1;
then
A11: ua in UU\/{ua} by XBOOLE_0:def 3;
ua = 1_K*u+a*u by A9,VECTSP_1:def 17
.= (1.K+a)*u by VECTSP_1:def 15;
then (1.K+a)" *ua =u by A10,VECTSP_1:20;
then u in Lin (UU\/{ua}) by A11,VECTSP_4:21,VECTSP_7:8;
hence thesis by A5,STRUCT_0:def 5;
end;
suppose
u=0.V;
then x in Lin(UU\/{ua}) by A5,VECTSP_4:17;
hence thesis by STRUCT_0:def 5;
end;
end;
end;
suppose
x<>u;
then x in UU by A4,ZFMISC_1:56;
then x in UU\/{ua} by XBOOLE_0:def 3;
then x in Lin (UU\/{ua}) by VECTSP_7:8;
hence thesis by STRUCT_0:def 5;
end;
end;
then Lin U is Subspace of Lin (UU\/{ua}) by VECTSP_9:16;
then
A12: the carrier of Lin U c=the carrier of Lin (UU\/{ua}) by VECTSP_4:def 2;
Lin (UU\/{ua}) is Subspace of Lin U by A1,A2,Th13;
then the carrier of Lin (UU\/{ua})c= the carrier of Lin U by VECTSP_4:def 2;
then the carrier of Lin U =the carrier of Lin (UU\/ {ua}) by A12;
hence thesis by VECTSP_4:29;
end;
begin :: Selected properties of the operator ^^
definition
let D be non empty set;
let n,m,k be Nat;
let A be Matrix of n,m,D;
let B be Matrix of n,k,D;
redefine func A^^B -> Matrix of n,width A+width B,D;
coherence
proof
reconsider N=n as Element of NAT by ORDINAL1:def 12;
set AB=A^^B;
A1: dom B = Seg len B by FINSEQ_1:def 3
.= Seg n by MATRIX_0:def 2;
dom A = Seg len A by FINSEQ_1:def 3
.= Seg n by MATRIX_0:def 2;
then
A2: dom AB = Seg n/\Seg n by A1,PRE_POLY:def 4
.= Seg n;
then
A3: len AB=N by FINSEQ_1:def 3;
per cases;
suppose
A4: N=0;
then AB={} by A2;
hence thesis by A4,MATRIX_0:13;
end;
suppose
A5: N>0;
then
A6: N in Seg N by FINSEQ_1:3;
then A.N=Line(A,N) & B.N=Line(B,N) by MATRIX_0:52;
then
A7: AB.N=Line(A,N)^Line(B,N) by A2,A6,PRE_POLY:def 4;
A8: len (Line(A,N)^Line(B,N))=width A+width B by CARD_1:def 7;
AB.N in rng AB by A2,A6,FUNCT_1:def 3;
then width AB=width A+width B by A3,A5,A7,A8,MATRIX_0:def 3;
hence thesis by A3,MATRIX_0:51;
end;
end;
end;
theorem Th15:
for D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k
,D for i st i in Seg n holds Line(A^^B,i)=Line(A,i)^Line(B,i)
proof
let D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k,D;
set AB=A^^B;
A1: len AB=n & dom AB=Seg len AB by FINSEQ_1:def 3,MATRIX_0:def 2;
let i such that
A2: i in Seg n;
Line(A,i)=A.i & Line(B,i)=B.i by A2,MATRIX_0:52;
hence Line(A,i)^Line(B,i) = AB.i by A2,A1,PRE_POLY:def 4
.= Line(AB,i) by A2,MATRIX_0:52;
end;
theorem Th16:
for D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k
,D for i st i in Seg width A holds Col(A^^B,i) = Col(A,i)
proof
let D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k,D;
let i such that
A1: i in Seg width A;
set AB=A^^B;
A2: len AB=n by MATRIX_0:def 2;
A3: len A=n by MATRIX_0:def 2;
now
let j such that
A4: j in Seg n;
n<>0 by A4;
then width AB=width A+width B by MATRIX_0:23;
then width A<=width AB by NAT_1:11;
then
A5: Seg width A c= Seg width AB by FINSEQ_1:5;
A6: dom A=Seg n by A3,FINSEQ_1:def 3;
A7: dom Line(A,j)=Seg width A by FINSEQ_2:124;
dom AB=Seg n by A2,FINSEQ_1:def 3;
hence Col(AB,i).j = AB*(j,i) by A4,MATRIX_0:def 8
.= Line(AB,j).i by A1,A5,MATRIX_0:def 7
.= (Line(A,j)^Line(B,j)).i by A4,Th15
.= Line(A,j).i by A1,A7,FINSEQ_1:def 7
.= A*(j,i) by A1,MATRIX_0:def 7
.= Col(A,i).j by A4,A6,MATRIX_0:def 8;
end;
hence thesis by A3,A2,FINSEQ_2:119;
end;
theorem Th17:
for D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k
,D for i st i in Seg width B holds Col(A^^B,width A+i) = Col(B,i)
proof
let D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k,D;
let i such that
A1: i in Seg width B;
set AB=A^^B;
A2: len AB=n by MATRIX_0:def 2;
A3: len B=n by MATRIX_0:def 2;
now
A4: dom B=Seg n by A3,FINSEQ_1:def 3;
let j such that
A5: j in Seg n;
n<>0 by A5;
then width AB=width A+width B by MATRIX_0:23;
then
A6: width A+i in Seg width AB by A1,FINSEQ_1:60;
A7: dom Line(B,j)=Seg width B & len Line(A,j)=width A by CARD_1:def 7
,FINSEQ_2:124;
dom AB=Seg n by A2,FINSEQ_1:def 3;
hence Col(AB,width A+i).j = AB*(j,width A+i) by A5,MATRIX_0:def 8
.= Line(AB,j).(width A+i) by A6,MATRIX_0:def 7
.= (Line(A,j)^Line(B,j)).(width A+i) by A5,Th15
.= Line(B,j).i by A1,A7,FINSEQ_1:def 7
.= B*(j,i) by A1,MATRIX_0:def 7
.= Col(B,i).j by A5,A4,MATRIX_0:def 8;
end;
hence thesis by A3,A2,FINSEQ_2:119;
end;
theorem Th18:
for D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k
,D for pA,pB be FinSequence of D st len pA = width A & len pB = width B holds
ReplaceLine(A^^B,i,pA^pB) = ReplaceLine(A,i,pA) ^^ ReplaceLine(B,i,pB)
proof
let D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k,D;
let pA,pB be FinSequence of D such that
A1: len pA = width A and
A2: len pB = width B;
set RB=RLine(B,i,pB);
set RA=RLine(A,i,pA);
set AB=A^^B;
set RAB=RLine(AB,i,pA^pB);
set Rab=RA^^RB;
A3: now
pA is Element of (width A)-tuples_on D & pB is Element of (width B)
-tuples_on D by A1,A2,FINSEQ_2:92;
then pA^pB is Tuple of width A+width B, D;
then pA^pB is Element of (width A+width B)-tuples_on D by FINSEQ_2:131;
then
A4: len (pA^pB)=width A+width B by CARD_1:def 7;
let j such that
A5: 1<=j & j<=n;
A6: j in Seg n by A5;
A7: width AB=width A+width B by A5,MATRIX_0:23;
A8: now
per cases;
suppose
A9: i=j;
then
A10: Line(RB,j)=pB by A2,A6,MATRIX11:28;
Line(RAB,j)=pA^pB & Line(RA,j)=pA by A1,A6,A4,A7,A9,MATRIX11:28;
hence Line(RAB,j)=Line(Rab,j) by A6,A10,Th15;
end;
suppose
A11: i<>j;
then
A12: Line(RB,j)=Line(B,j) by A6,MATRIX11:28;
Line(RAB,j)=Line(AB,j) & Line(RA,j)=Line(A,j) by A6,A11,MATRIX11:28;
hence Line(RAB,j) = Line(RA,j)^Line(RB,j) by A6,A12,Th15
.= Line(Rab,j) by A6,Th15;
end;
end;
thus RAB.j = Line(RAB,j) by A6,MATRIX_0:52
.= Rab.j by A6,A8,MATRIX_0:52;
end;
len RAB=n & len Rab=n by MATRIX_0:def 2;
hence thesis by A3;
end;
theorem Th19:
for D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k
,D holds Segm(A^^B,Seg n,Seg width A) = A & Segm(A^^B,Seg n,Seg (width A+width
B)\Seg width A) = B
proof
let D be non empty set, A be Matrix of n,m,D, B be Matrix of n,k,D;
set AB=A^^B;
A1: card Seg n=n by FINSEQ_1:57;
A2: len A=n by MATRIX_0:def 2;
then reconsider A9=A as Matrix of n,width A,D by MATRIX_0:51;
set S1=Segm(AB,Seg n,Seg width A);
A3: card Seg width A=width A by FINSEQ_1:57;
A4: len AB=n by MATRIX_0:def 2;
now
let i,j such that
A5: [i,j] in Indices A9;
reconsider I=i,J=j as Element of NAT by ORDINAL1:def 12;
A6: dom A=Seg n by A2,FINSEQ_1:def 3;
n<>0 by A5,MATRIX_0:22;
then Indices A9=[:Seg n,Seg width A:] by MATRIX_0:23;
then
A7: I in Seg n by A5,ZFMISC_1:87;
A8: J in Seg width A by A5,ZFMISC_1:87;
then
A9: J = (idseq width A).J by FINSEQ_2:49
.= Sgm (Seg (width A)).J by FINSEQ_3:48;
dom S1 = Seg len S1 by FINSEQ_1:def 3
.= Seg n by A1,MATRIX_0:def 2;
hence S1*(i,j) = Col(S1,J).I by A7,MATRIX_0:def 8
.= Col(AB,Sgm (Seg (width A)).J).I by A4,A3,A8,MATRIX13:50
.= Col(A,J).I by A8,A9,Th16
.= A*(i,j) by A7,A6,MATRIX_0:def 8;
end;
hence A=S1 by A1,A3,MATRIX_0:27;
set w=width A+width B;
set SS=Seg w\Seg width A;
set S2=Segm(AB,Seg n,SS);
A10: len B=n by MATRIX_0:def 2;
then reconsider B9=B as Matrix of n,width B,D by MATRIX_0:51;
width A <= w by NAT_1:11;
then card Seg (width A+width B)=width A+width B & Seg width A c= Seg w by
FINSEQ_1:5,57;
then
A11: card SS = w-width A by A3,CARD_2:44
.= width B;
now
A12: dom B = Seg n by A10,FINSEQ_1:def 3;
let i,j such that
A13: [i,j] in Indices B9;
reconsider I=i,J=j as Element of NAT by ORDINAL1:def 12;
A14: J in Seg width B by A13,ZFMISC_1:87;
n<>0 by A13,MATRIX_0:22;
then Indices B9=[:Seg n,Seg width B:] by MATRIX_0:23;
then
A15: I in Seg n by A13,ZFMISC_1:87;
dom S2 = Seg len S2 by FINSEQ_1:def 3
.= Seg n by A1,MATRIX_0:def 2;
hence S2*(i,j) = Col(S2,J).I by A15,MATRIX_0:def 8
.= Col(AB,Sgm SS.J).I by A4,A11,A14,MATRIX13:50
.= Col(AB,width A+J).I by A14,Th8
.= Col(B,J).I by A14,Th17
.= B9*(i,j) by A15,A12,MATRIX_0:def 8;
end;
hence thesis by A1,A11,MATRIX_0:27;
end;
theorem Th20:
for A,B be Matrix of K st len A = len B holds the_rank_of A <=
the_rank_of (A^^B) & the_rank_of B <= the_rank_of (A^^B)
proof
let A,B be Matrix of K such that
A1: len A = len B;
set L=len A;
reconsider B9=B as Matrix of L,width B,K by A1,MATRIX_0:51;
reconsider A9=A as Matrix of L,width A,K by MATRIX_0:51;
set AB=A9^^B9;
per cases;
suppose
L=0;
hence thesis by A1,MATRIX13:74;
end;
suppose
A2: L>0;
A3: Segm(AB,Seg L,Seg width A) = A by Th19;
A4: Indices AB=[:Seg L,Seg (width A+ width B):] by A2,MATRIX_0:23;
A5: width AB=width A+width B by A2,MATRIX_0:23;
then width A <=width AB by NAT_1:11;
then Seg width A c=Seg width AB by FINSEQ_1:5;
then
A6: [:Seg L,Seg width A:] c= Indices AB by A5,A4,ZFMISC_1:95;
Seg width AB\Seg width A c=Seg width AB by XBOOLE_1:36;
then
A7: [:Seg L,Seg width AB\Seg width A:] c= Indices AB by A5,A4,ZFMISC_1:95;
Segm(AB,Seg L,Seg width AB\Seg width A) = B by A5,Th19;
hence thesis by A3,A6,A7,MATRIX13:79;
end;
end;
theorem
for A,B be Matrix of K st len A = len B & len A = the_rank_of A holds
the_rank_of A = the_rank_of (A^^B)
proof
let A,B be Matrix of K such that
A1: len A = len B and
A2: len A =the_rank_of A;
set L=len A;
reconsider B9=B as Matrix of L,width B,K by A1,MATRIX_0:51;
reconsider A9=A as Matrix of L,width A,K by MATRIX_0:51;
A3: the_rank_of (A9^^B9)<=len (A9^^B9) & len (A9^^B9)=L by MATRIX13:74
,MATRIX_0:def 2;
the_rank_of (A9^^B9)>=L by A1,A2,Th20;
hence thesis by A2,A3,XXREAL_0:1;
end;
theorem Th22:
for A,B be Matrix of K st len A = len B & width A = 0 holds A ^^
B = B & B ^^ A = B
proof
let A,B be Matrix of K such that
A1: len A = len B and
A2: width A =0;
A3: Seg width A = {} by A2;
set L=len A;
reconsider B9=B as Matrix of L,width B,K by A1,MATRIX_0:51;
reconsider A9=A as Matrix of L,width A,K by MATRIX_0:51;
set AB=A9^^B9;
set BA=B9^^A9;
per cases;
suppose
A4: L=0;
then len BA=0 by MATRIX_0:def 2;
then
A5: BA={};
len AB=0 by A4,MATRIX_0:def 2;
then AB={};
hence thesis by A1,A4,A5;
end;
suppose
A6: L>0;
then width AB=width B & len AB=L by A2,MATRIX_0:23;
hence A^^B = Segm(AB,Seg L,Seg width B\Seg width A) by A3,MATRIX13:46
.= B by A2,Th19;
width BA=width B & len BA=L by A2,A6,MATRIX_0:23;
hence B^^A = Segm(BA,Seg L,Seg width B) by MATRIX13:46
.=B by Th19;
end;
end;
theorem Th23:
for A,B be Matrix of K st B = 0.(K,len A,m) holds the_rank_of A
= the_rank_of (A^^B)
proof
let A,B be Matrix of K such that
A1: B = 0.(K,len A,m);
A2: len B = len A by A1,MATRIX_0:def 2;
set L=len A;
reconsider B9=B as Matrix of L,width B,K by A2,MATRIX_0:51;
reconsider A9=A as Matrix of L,width A,K by MATRIX_0:51;
set AB=A9^^B9;
per cases;
suppose
width B=0;
hence thesis by A2,Th22;
end;
suppose
width B>0;
then
A3: L>0 by A2,MATRIX_0:def 3;
then
A4: width AB=width A+width B by MATRIX_0:23;
A5: len AB=L by A3,MATRIX_0:23;
A6: now
set L0=len AB |-> 0.K;
let i such that
A7: i in Seg width AB \ Seg width A;
A8: i in Seg width AB by A7,XBOOLE_0:def 5;
not i in Seg width A by A7,XBOOLE_0:def 5;
then
A9: i<1 or i>width A;
then reconsider n=i-width A as Element of NAT by A8,FINSEQ_1:1,NAT_1:21;
A10: i=n+width A;
n<>0 by A8,A9,FINSEQ_1:1;
then
A11: n in Seg width B by A4,A8,A10,FINSEQ_1:61;
A12: now
let i such that
A13: 1<=i & i<=L;
A14: i in Seg L by A13;
then
A15: L0.i=0.K by A5,FINSEQ_2:57;
Seg L=dom B by A2,FINSEQ_1:def 3;
then Col(B9,n).i=B9*(i,n) & [i,n] in Indices B by A11,A14,
MATRIX_0:def 8,ZFMISC_1:87;
hence Col(B9,n).i=L0.i by A1,A15,MATRIX_3:1;
end;
A16: len Col(B,n)=L & len L0=L by A2,A5,CARD_1:def 7;
Col(AB,i)=Col(B9,n) by A10,A11,Th17;
hence Col(AB,i) = len AB |-> 0.K by A16,A12;
end;
width A <= width AB by A4,NAT_1:11;
then Seg width A c= Seg width AB by FINSEQ_1:5;
hence the_rank_of (A^^B) = the_rank_of Segm(AB,Seg len AB,Seg width A) by
A6,Th12
.= the_rank_of A by A5,Th19;
end;
end;
theorem Th24:
for A,B be Matrix of K st the_rank_of A = the_rank_of (A^^B) &
len A = len B for N st N c= dom A & for i st i in N holds Line(A,i)=width A|->
0.K holds for i st i in N holds Line(B,i) = width B|->0.K
proof
let A,B be Matrix of K such that
A1: the_rank_of A = the_rank_of (A^^B) and
A2: len A=len B;
reconsider B9=B as Matrix of len A,width B,K by A2,MATRIX_0:51;
reconsider A9=A as Matrix of len A,width A,K by MATRIX_0:51;
set AB=A9^^B9;
let N such that
A3: N c= dom A and
A4: for i st i in N holds Line(A,i)=width A|->0.K;
let i such that
A5: i in N;
dom A <> {} by A3,A5;
then Seg len A <> {} by FINSEQ_1:def 3;
then
A6: len A<>0;
then
A7: width AB=width A+width B by MATRIX_0:23;
then width A <= width AB by NAT_1:11;
then
A8: Seg width A c= Seg width AB by FINSEQ_1:5;
A9: card Seg len A=len A by FINSEQ_1:57;
A10: Segm(AB,Seg len A,Seg width A)=A by Th19;
A11: dom A=Seg len A by FINSEQ_1:def 3;
A12: Sgm (Seg len A).i = (idseq len A).i by FINSEQ_3:48
.= i by A3,A5,A11,FINSEQ_2:49;
card Seg width A=width A by FINSEQ_1:57;
then
A13: card (Seg width A) |->0.K = Line(A,i) by A4,A5
.= Line(AB,i) * Sgm (Seg width A) by A3,A5,A11,A10,A8,A9,A12,MATRIX13:47;
assume Line(B,i)<>width B|->0.K;
then consider j such that
A14: j in Seg width B and
A15: Line(B,i).j<>(width B|->0.K).j by FINSEQ_2:119;
A16: len Line(A9,i)=width A9 & 1<=j by A14,FINSEQ_1:1,MATRIX_0:def 7;
len Line(B9,i)=width B9 by MATRIX_0:def 7;
then
A17: j<=len Line(B9,i) by A14,FINSEQ_1:1;
A18: j+width A in Seg width AB by A14,A7,FINSEQ_1:60;
then AB*(i,j+width A) = Line(AB,i).(j+width A) by MATRIX_0:def 7
.= (Line(A9,i)^Line(B9,i)).(j+width A) by A3,A5,A11,Th15
.= Line(B9,i).j by A16,A17,FINSEQ_1:65;
then
A19: AB*(i,j+width A)<> 0.K by A14,A15,FINSEQ_2:57;
consider P,Q be without_zero finite Subset of NAT such that
A20: [:P,Q:] c= Indices A9 and
A21: card P = card Q and
A22: card P = the_rank_of A9 and
A23: Det EqSegm(A9,P,Q)<>0.K by MATRIX13:def 4;
P={} iff Q={} by A21;
then consider P2,Q2 be without_zero finite Subset of NAT such that
A24: P2 c= Seg len A and
A25: Q2 c= Seg width A and
A26: P2 = Sgm Seg len A.:P and
Q2=Sgm Seg width A.:Q and
card P2=card P and
card Q2=card Q and
A27: Segm(A,P,Q) = Segm(AB,P2,Q2) by A20,A10,MATRIX13:57;
A28: Segm(AB,P2,Q2)=EqSegm(A,P,Q) by A21,A27,MATRIX13:def 3;
A29: dom AB=Seg len AB & len AB=len A by A6,FINSEQ_1:def 3,MATRIX_0:23;
then
A30: [:P2,Seg width A:] c= Indices AB by A24,A8,ZFMISC_1:96;
j>=1 by A14,FINSEQ_1:1;
then j+width A >=1+width A by XREAL_1:6;
then j+width A >width A by NAT_1:13;
then not j+width A in Q2 by A25,FINSEQ_1:1;
then
A31: j+width A in Seg width AB\Q2 by A18,XBOOLE_0:def 5;
not i in P2
proof
A32: Line(A,i) = width A|->0.K by A4,A5
.= 0.K*Line(A,i) by FVSUM_1:58;
A33: Sgm Seg len A = idseq len A by FINSEQ_3:48
.= id (Seg len A);
A34: P c= Seg len A by A20,A21,MATRIX13:67;
then
A35: rng Sgm P=P by FINSEQ_1:def 13;
assume i in P2;
then i in P by A26,A33,A34,FUNCT_1:92;
then consider x being object such that
A36: x in dom Sgm P and
A37: Sgm P.x=i by A35,FUNCT_1:def 3;
reconsider x as Element of NAT by A36;
A38: Segm(A,P,Q)=EqSegm(A,P,Q) by A21,MATRIX13:def 3;
A39: Q c= Seg width A by A20,A21,MATRIX13:67;
then
A40: rng Sgm Q=Q by FINSEQ_1:def 13;
A41: dom Sgm P=Seg card P by A34,FINSEQ_3:40;
then dom Line(A,i)=Seg width A & Line(Segm(A,P,Q),x)=Line(A,i)* Sgm Q by
A39,A36,A37,FINSEQ_2:124,MATRIX13:47;
then Line(Segm(A,P,Q),x)=0.K * Line(Segm(A,P,Q),x) by A39,A40,A32,
MATRIX13:87;
then
0.K*Det EqSegm(A,P,Q) = Det RLine(EqSegm(A,P,Q),x,Line(EqSegm(A,P,Q),
x)) by A41,A36,A38,MATRIX11:35
.= Det EqSegm(A,P,Q) by MATRIX11:30;
hence thesis by A23;
end;
then i in dom AB\P2 by A3,A5,A11,A29,XBOOLE_0:def 5;
then card P > the_rank_of Segm(AB,P2,Q2) by A1,A19,A22,A25,A30,A31,A13,Th10;
hence thesis by A23,A28,MATRIX13:83;
end;
begin ::Basic properties of two transformations which
::transform finite sequences to matrices
reserve D for non empty set,
bD for FinSequence of D,
b,f,g for FinSequence of K,
MD for Matrix of D;
definition
let D be non empty set;
let b be FinSequence of D;
func LineVec2Mx b -> Matrix of 1,len b,D equals
<*b*>;
coherence;
func ColVec2Mx b -> Matrix of len b,1,D equals
<*b*>@;
coherence
proof
set B=<*b*>;
A1: len B=1 by MATRIX_0:23;
A2: width B=len b by MATRIX_0:23;
A3: len (B@)=width B by MATRIX_0:def 6;
per cases;
suppose
A4: len b=0;
then B@={} by A3,MATRIX_0:23;
hence thesis by A4,MATRIX_0:13;
end;
suppose
len b>0;
then width (B@)=len B by A2,MATRIX_0:54;
hence thesis by A1,A2,A3,MATRIX_0:51;
end;
end;
end;
theorem Th25:
MD = LineVec2Mx bD iff Line(MD,1) = bD & len MD = 1
proof
thus MD = LineVec2Mx bD implies Line(MD,1) = bD & len MD = 1
proof
1 in Seg 1;
then
A1: Line(LineVec2Mx bD,1)=(LineVec2Mx bD).1 by MATRIX_0:52;
assume MD = LineVec2Mx bD;
hence thesis by A1,FINSEQ_1:40;
end;
assume that
A2: Line(MD,1) = bD and
A3: len MD = 1;
reconsider md=MD as Matrix of 1,width MD,D by A3,MATRIX_0:51;
1 in Seg 1;
then md.1=bD by A2,MATRIX_0:52;
hence thesis by A3,FINSEQ_1:40;
end;
theorem Th26:
( len MD <> 0 or len bD <> 0 ) implies ( MD = ColVec2Mx bD iff
Col(MD,1) = bD & width MD = 1 )
proof
assume
A1: len MD <> 0 or len bD <> 0;
thus MD = ColVec2Mx bD implies Col(MD,1) = bD & width MD = 1
proof
len (LineVec2Mx bD)=1 by Th25;
then
A2: dom (LineVec2Mx bD)=Seg 1 by FINSEQ_1:def 3;
assume
A3: MD = ColVec2Mx bD;
1 in Seg 1;
hence Col(MD,1) = Line(LineVec2Mx bD,1) by A3,A2,MATRIX_0:58
.= bD by Th25;
len MD=len bD by A3,MATRIX_0:def 2;
hence thesis by A1,A3,MATRIX_0:23;
end;
assume that
A4: Col(MD,1) = bD and
A5: width MD = 1;
A6: len MD>0 by A1,A4,MATRIX_0:def 8;
A7: len (MD@)=1 by A5,MATRIX_0:def 6;
1 in Seg 1;
then Line(MD@,1)=bD by A4,A5,MATRIX_0:59;
then (LineVec2Mx bD)@ = (MD@)@ by A7,Th25
.= MD by A5,A6,MATRIX_0:57;
hence thesis;
end;
theorem
len f = len g implies (LineVec2Mx f) + (LineVec2Mx g) = LineVec2Mx (f + g)
proof
set Lf=LineVec2Mx f;
set Lg=LineVec2Mx g;
A1: len Lf=1 by CARD_1:def 7;
assume
A2: len f = len g;
then reconsider F=f,G=g as Element of (len f)-tuples_on the carrier of K by
FINSEQ_2:92;
A3: width Lg=len f by A2,MATRIX_0:23;
set FG=F+G;
set Lfg=LineVec2Mx FG;
A4: len FG=len f & len Lfg=1 by CARD_1:def 7;
A5: width Lfg=len FG by MATRIX_0:23;
A6: len (Lf+Lg)=len Lf by MATRIX_3:def 3;
A7: width (Lf+Lg)=width Lf by MATRIX_3:def 3;
A8: width Lf=len f by MATRIX_0:23;
per cases;
suppose
len f=0;
then Seg len f = {};
then for i,j st [i,j] in Indices (Lf+Lg) holds (Lf+Lg)*(i,j) = Lfg*(i,j)
by A7,A8,ZFMISC_1:90;
hence thesis by A4,A1,A6,A7,A8,A5,MATRIX_0:21;
end;
suppose
A9: len f>0;
A10: dom Lf=Seg 1 & 1 in Seg 1 by A1,FINSEQ_1:def 3;
len f in Seg len f by A9,FINSEQ_1:3;
then [1,len f] in Indices Lf by A8,A10,ZFMISC_1:87;
then Line(Lf+Lg,1) = Line(Lf,1)+Line(Lg,1) by A3,MATRIX_0:23,MATRIX_4:59
.= f+Line(Lg,1) by Th25
.= f+g by Th25;
hence thesis by A1,A6,Th25;
end;
end;
theorem Th28:
len f = len g implies (ColVec2Mx f) + (ColVec2Mx g) = ColVec2Mx (f + g)
proof
set Cf=ColVec2Mx f;
set Cg=ColVec2Mx g;
A1: len Cf=len f by MATRIX_0:def 2;
assume
A2: len f = len g;
then reconsider F=f,G=g as Element of (len f)-tuples_on the carrier of K by
FINSEQ_2:92;
A3: len Cg=len f by A2,MATRIX_0:def 2;
set FG=F+G;
set Cfg=ColVec2Mx FG;
A4: len Cfg=len FG by MATRIX_0:def 2;
A5: len FG=len f & width (Cf+Cg)=width Cf by CARD_1:def 7,MATRIX_3:def 3;
A6: len (Cf+Cg)=len Cf by MATRIX_3:def 3;
per cases;
suppose
A7: len f=0;
then Cf+Cg={} by A6,MATRIX_0:def 2;
hence thesis by A4,A7;
end;
suppose
A8: len f>0;
A9: dom Cf=Seg len f & 1 in Seg 1 by A1,FINSEQ_1:def 3;
A10: width Cf= 1 by A8,MATRIX_0:23;
len f in Seg len f by A8,FINSEQ_1:3;
then [len f,1] in Indices Cf by A9,A10,ZFMISC_1:87;
then Col(Cf+Cg,1) = Col(Cf,1)+Col(Cg,1) by A1,A3,MATRIX_4:60
.= f+Col(Cg,1) by A8,Th26
.= f+g by A2,A8,Th26;
hence thesis by A5,A8,A10,Th26;
end;
end;
theorem
a * LineVec2Mx f =LineVec2Mx (a*f)
proof
A1: len (a*LineVec2Mx f)=len (LineVec2Mx f) by MATRIX_3:def 5;
A2: len (LineVec2Mx f)=1 by MATRIX_0:def 2;
then Line(a * LineVec2Mx f,1) = a*Line(LineVec2Mx f,1) by MATRIXR1:20
.= a*f by Th25;
hence thesis by A2,A1,Th25;
end;
theorem Th30:
a * ColVec2Mx f = ColVec2Mx (a*f)
proof
A1: len f=len (a*f) by MATRIXR1:16;
per cases;
suppose
A2: len f=0;
len (ColVec2Mx f)=len (a*ColVec2Mx f) by MATRIX_3:def 5;
then
A3: a * ColVec2Mx f = {} by A2,MATRIX_0:def 2;
len (ColVec2Mx (a*f))= 0 by A1,A2,MATRIX_0:def 2;
hence thesis by A3;
end;
suppose
A4: len f>0;
A5: width (a*ColVec2Mx f)=width ColVec2Mx f by MATRIX_3:def 5;
A6: width ColVec2Mx f=1 by A4,MATRIX_0:23;
then Col(a * ColVec2Mx f,1) = a*Col(ColVec2Mx f,1) by MATRIXR1:19
.= a*f by A4,Th26;
hence thesis by A1,A4,A6,A5,Th26;
end;
end;
theorem
LineVec2Mx (k|->0.K) = 0.(K,1,k)
proof
card(k|->0.K) = k by CARD_1:def 7;
then reconsider L=LineVec2Mx (k|->0.K) as Matrix of 1,k,K;
set Z=0.(K,1,k);
now
A1: width L=k by MATRIX_0:23;
A2: dom L = Seg len L & len L=1 by FINSEQ_1:def 3,MATRIX_0:def 2;
let i,j such that
A3: [i,j] in Indices L;
A4: j in Seg width L by A3,ZFMISC_1:87;
i in dom L by A3,ZFMISC_1:87;
then
A5: i=1 by A2,FINSEQ_1:2,TARSKI:def 1;
Indices Z=Indices L by MATRIX_0:26;
hence Z*(i,j) = 0.K by A3,MATRIX_3:1
.= (k|->0.K).j by A4,A1,FINSEQ_2:57
.= Line(L,i).j by A5,Th25
.= L*(i,j) by A4,MATRIX_0:def 7;
end;
hence thesis by MATRIX_0:27;
end;
theorem Th32:
ColVec2Mx (k|->0.K) = 0.(K,k,1)
proof
card(k|->0.K) = k by CARD_1:def 7;
then reconsider C=ColVec2Mx (k|->0.K) as Matrix of k,1,K;
set Z=0.(K,k,1);
now
A1: len (k|->0.K)=k by CARD_1:def 7;
let i,j such that
A2: [i,j] in Indices C;
A3: i in dom C by A2,ZFMISC_1:87;
A4: j in Seg width C by A2,ZFMISC_1:87;
A5: dom C=Seg len C & len C=len (k|->0.K) by FINSEQ_1:def 3,MATRIX_0:def 2;
then width C=1 by A3,A1,Th26;
then
A6: j=1 by A4,FINSEQ_1:2,TARSKI:def 1;
Indices Z=Indices C by MATRIX_0:26;
hence Z*(i,j) = 0.K by A2,MATRIX_3:1
.= (k|->0.K).i by A3,A5,A1,FINSEQ_2:57
.= Col(C,j).i by A3,A5,A1,A6,Th26
.= C*(i,j) by A3,MATRIX_0:def 8;
end;
hence thesis by MATRIX_0:27;
end;
begin :: Basis properties of the Solution of Linear Equations
definition
let K;
let A,B;
func Solutions_of(A,B) -> set equals
{X : len X = width A & width X = width B & A *
X = B};
coherence;
end;
theorem Th33:
Solutions_of(A,B) is non empty implies len A = len B
proof
assume Solutions_of(A,B) is non empty;
then consider x being object such that
A1: x in Solutions_of(A,B);
ex X st X=x & len X= width A & width X = width B & A * X = B by A1;
hence thesis by MATRIX_3:def 4;
end;
theorem
X in Solutions_of(A,B) & i in Seg width X & Col(X,i) = len X |-> 0.K
implies Col(B,i) = len B |-> 0.K
proof
assume that
A1: X in Solutions_of(A,B) and
A2: i in Seg width X and
A3: Col(X,i) = len X |-> 0.K;
set LB0=len B |-> 0.K;
consider X1 such that
A4: X = X1 and
A5: len X1 = width A and
A6: width X1 = width B and
A7: A * X1 = B by A1;
A8: now
let k such that
A9: 1 <=k & k <= len B;
A10: k in Seg len B by A9;
Indices B=[:Seg len B,Seg width B:] by FINSEQ_1:def 3;
then [k,i] in Indices B by A2,A4,A6,A10,ZFMISC_1:87;
then
A11: B*(k,i) = Line(A,k) "*" Col(X1,i) by A5,A7,MATRIX_3:def 4
.= Sum(0.K*Line(A,k)) by A3,A4,A5,FVSUM_1:66
.= 0.K* Sum(Line(A,k)) by FVSUM_1:73
.= 0.K
.= LB0.k by A10,FINSEQ_2:57;
k in dom B by A10,FINSEQ_1:def 3;
hence (Col(B,i)).k=LB0.k by A11,MATRIX_0:def 8;
end;
len Col(B,i) = len B & len LB0 = len B by CARD_1:def 7;
hence thesis by A8;
end;
theorem Th35:
X in Solutions_of(A,B) implies a*X in Solutions_of(A,a*B) & X in
Solutions_of(a*A,a*B)
proof
A1: width (a*B) = width B & width (a*A)=width A by MATRIX_3:def 5;
assume X in Solutions_of(A,B);
then consider X1 such that
A2: X = X1 & len X1 = width A and
A3: width X1 = width B & A * X1 = B;
A4: len (a*X) = width A & width (a*X) = width X1 by A2,MATRIX_3:def 5;
A*(a*X)=a*(A*X) & (a*A)*X=a*(A*X) by A2,Th1,MATRIXR1:22;
hence thesis by A2,A3,A4,A1;
end;
theorem
a <> 0.K implies Solutions_of(A,B) = Solutions_of(a*A,a*B)
proof
assume
A1: a<>0.K;
thus Solutions_of(A,B) c= Solutions_of(a*A,a*B)
proof
let x be object such that
A2: x in Solutions_of(A,B);
ex X st x=X & len X = width A & width X = width B & A * X = B by A2;
hence thesis by A2,Th35;
end;
A3: a"*(a*A) = (a"*a)*A by Th2
.= 1_K *A by A1,VECTSP_1:def 10
.= A by Th2;
A4: a"*(a*B) = (a"*a)*B by Th2
.= 1_K*B by A1,VECTSP_1:def 10
.= B by Th2;
let x be object such that
A5: x in Solutions_of(a*A,a*B);
ex X st x=X & len X = width (a*A) & width X = width (a* B) & (a*A) * X =
(a*B) by A5;
hence thesis by A5,A3,A4,Th35;
end;
Lm3: for D be non empty set for A,B be Matrix of D st len A = len B & width A
= 0 & width B = 0 holds A = B
proof
let D be non empty set;
let A,B be Matrix of D such that
A1: len A = len B and
A2: width A = 0 and
A3: width B = 0;
Seg width A = {} by A2;
then Indices A = {} by ZFMISC_1:90;
then
for i,j st [i,j] in Indices A holds A*(i,j) = B*(i,j);
hence thesis by A1,A2,A3,MATRIX_0:21;
end;
theorem Th37:
X1 in Solutions_of(A,B1) & X2 in Solutions_of(A,B2) & width B1 =
width B2 implies X1 + X2 in Solutions_of(A,B1 + B2)
proof
assume that
A1: X1 in Solutions_of(A,B1) and
A2: X2 in Solutions_of(A,B2) and
A3: width B1 = width B2;
A4: ex Y1 be Matrix of K st Y1 = X1 & len Y1 = width A & width Y1 = width B1
& A * Y1 = B1 by A1;
A5: width (X1+X2) = width X1 by MATRIX_3:def 3;
A6: len (X1+X2) = len X1 by MATRIX_3:def 3;
A7: ex Y2 be Matrix of K st Y2 = X2 & len Y2 = width A & width Y2 = width B2
& A * Y2 = B2 by A2;
A8: now
per cases;
suppose
A9: len A=0;
then len (A*X1) = 0 by A4,MATRIX_3:def 4;
then
A10: len (A*X1 + A*X2) = 0 by MATRIX_3:def 3;
len (A*(X1+X2)) = 0 by A4,A6,A9,MATRIX_3:def 4;
hence A*(X1+X2) = A*X1 + A*X2 by A10;
end;
suppose
len X1=0;
then width (A*X1)=0 by A4,MATRIX_0:def 3;
then
A11: width (A*X1 + A*X2)=0 & width (A*(X1+X2)) = 0 by A4,A6,A5,MATRIX_3:def 3
,def 4;
len (A*X1)=len A by A4,MATRIX_3:def 4;
then
A12: len (A*X1+A*X2)=len A by MATRIX_3:def 3;
len (A*(X1+X2))= len A by A4,A6,MATRIX_3:def 4;
hence A*(X1+X2)=A*X1+A*X2 by A12,A11,Lm3;
end;
suppose
len A>0 & len X1>0;
hence A*(X1+X2)=A*X1 + A*X2 by A3,A4,A7,MATRIX_4:62;
end;
end;
width (B1+B2) = width B1 by MATRIX_3:def 3;
hence thesis by A4,A7,A6,A5,A8;
end;
theorem Th38:
X in Solutions_of(A9,B9) implies X in Solutions_of(RLine(A9,i,a*
Line(A9,i)),RLine(B9,i,a*Line(B9,i)))
proof
set LA=Line(A9,i);
set LB=Line(B9,i);
set RA=RLine(A9,i,a*LA);
set RB=RLine(B9,i,a*LB);
A1: Indices RB=Indices B9 by MATRIX_0:26;
A2: len (a*LB)=len LB & len LB=width B9 by CARD_1:def 7,MATRIXR1:16;
then
A3: width RB=width B9 by MATRIX11:def 3;
A4: len (a*LA)=len LA & len LA= width A9 by CARD_1:def 7,MATRIXR1:16;
then
A5: len RA=len A9 by MATRIX11:def 3;
assume
A6: X in Solutions_of(A9,B9);
then consider X1 be Matrix of K such that
A7: X = X1 and
A8: len X1 = width A9 and
A9: width X1 = width B9 and
A10: A9 * X1 = B9;
set RX=RA*X1;
A11: width RA=width A9 by A4,MATRIX11:def 3;
then
A12: len RX=len RA & width RX=width X1 by A8,MATRIX_3:def 4;
A13: len A9=len B9 by A6,Th33;
then dom B9=Seg len RA by A5,FINSEQ_1:def 3;
then
A14: Indices RX=Indices B9 by A9,A12,FINSEQ_1:def 3;
A15: now
len B9=m by MATRIX_0:def 2;
then
A16: dom B9=Seg m by FINSEQ_1:def 3;
let j,k such that
A17: [j,k] in Indices RB;
A18: j in dom B9 by A1,A17,ZFMISC_1:87;
A19: k in Seg width B9 by A1,A17,ZFMISC_1:87;
then B9*(i,k)=LB.k by MATRIX_0:def 7;
then reconsider LBk=LB.k as Element of K;
A20: B9*(j,k)= Line(A9,j) "*" Col(X1,k) by A8,A10,A1,A17,MATRIX_3:def 4;
now
per cases;
suppose
A21: j=i;
then Line(RA,i)=a*LA by A4,A18,A16,MATRIX11:28;
hence RX*(j,k) = (a*LA)"*"Col(X1,k) by A8,A11,A14,A1,A17,A21,
MATRIX_3:def 4
.= Sum(a*mlt(LA,Col(X1,k))) by A8,FVSUM_1:68
.= a*Sum(mlt(LA,Col(X1,k))) by FVSUM_1:73
.= a*LBk by A19,A20,A21,MATRIX_0:def 7
.= (a*LB).k by A19,FVSUM_1:51
.= RB*(j,k) by A2,A1,A17,A21,MATRIX11:def 3;
end;
suppose
A22: j<>i;
then Line(RA,j)=Line(A9,j) by A18,A16,MATRIX11:28;
hence RX*(j,k) = Line(A9,j)"*"Col(X1,k) by A8,A11,A14,A1,A17,
MATRIX_3:def 4
.= B9*(j,k) by A8,A10,A1,A17,MATRIX_3:def 4
.= RB*(j,k) by A2,A1,A17,A22,MATRIX11:def 3;
end;
end;
hence RB*(j,k) = RX*(j,k);
end;
len RB=len B9 by A2,MATRIX11:def 3;
then RX=RB by A9,A13,A5,A3,A12,A15,MATRIX_0:21;
hence thesis by A7,A8,A9,A11,A3;
end;
Lm4: a <> 0.K implies Solutions_of(A9,B9) = Solutions_of(RLine(A9,i,a*Line(A9,
i)),RLine(B9,i,a*Line(B9,i)))
proof
assume
A1: a <> 0.K;
set RB=RLine(B9,i,a*Line(B9,i));
set RA=RLine(A9,i,a*Line(A9,i));
thus Solutions_of(A9,B9) c= Solutions_of(RA,RB)
proof
let x be object such that
A2: x in Solutions_of(A9,B9);
ex X st x=X & len X = width A9 & width X = width B9 & A9 * X = B9 by A2;
hence thesis by A2,Th38;
end;
let x be object such that
A3: x in Solutions_of(RA,RB);
per cases;
suppose
A4: not i in Seg m;
len A9=m by MATRIX_0:def 2;
then len B9=m & RA=A9 by A4,MATRIX13:40,MATRIX_0:def 2;
hence thesis by A3,A4,MATRIX13:40;
end;
suppose
A5: i in Seg m;
reconsider aLA=a*Line(A9,i),aLB=a*Line(B9,i),aLAR=a"*Line(RA,i), aLBR=a"*
Line(RB,i) as Element of (the carrier of K)* by FINSEQ_1:def 11;
set RRB=RLine(RB,i,a"*Line(RB,i));
set RRA=RLine(RA,i,a"*Line(RA,i));
A6: ex X st x=X & len X = width RA & width X = width RB & RA * X = RB by A3;
A7: len (a*Line(A9,i))=width A9 by CARD_1:def 7;
then
A8: a"*Line(RA,i) = a"*(a*Line(A9,i)) by A5,MATRIX11:28
.= (a"*a)*Line(A9,i) by FVSUM_1:54
.= 1_K * Line(A9,i) by A1,VECTSP_1:def 10
.= Line(A9,i) by FVSUM_1:57;
A9: len (a*Line(B9,i))=width B9 by CARD_1:def 7;
then
A10: a"*Line(RB,i) = a"*(a*Line(B9,i)) by A5,MATRIX11:28
.= (a"*a)*Line(B9,i) by FVSUM_1:54
.= 1_K * Line(B9,i) by A1,VECTSP_1:def 10
.= Line(B9,i) by FVSUM_1:57;
A11: width RB=width B9 by A9,MATRIX11:def 3;
A12: len (a"*Line(RB,i))=width RB by CARD_1:def 7;
then
A13: RRB = Replace(RB,i,aLBR) by MATRIX11:29
.= Replace(Replace(B9,i,aLB),i,aLBR) by A9,MATRIX11:29
.= Replace(B9,i,aLBR) by FUNCT_7:34
.= RLine(B9,i,Line(B9,i)) by A12,A11,A10,MATRIX11:29
.= B9 by MATRIX11:30;
A14: width RA=width A9 by A7,MATRIX11:def 3;
A15: len (a"*Line(RA,i))=width RA by CARD_1:def 7;
then RRA = Replace(RA,i,aLAR) by MATRIX11:29
.= Replace(Replace(A9,i,aLA),i,aLAR) by A7,MATRIX11:29
.= Replace(A9,i,aLAR) by FUNCT_7:34
.= RLine(A9,i,Line(A9,i)) by A15,A14,A8,MATRIX11:29
.= A9 by MATRIX11:30;
hence thesis by A3,A6,A13,Th38;
end;
end;
theorem Th39:
X in Solutions_of(A9,B9) & j in Seg m implies X in Solutions_of(
RLine(A9,i,Line(A9,i) + a*Line(A9,j)), RLine(B9,i,Line(B9,i) + a*Line(B9,j)))
proof
assume that
A1: X in Solutions_of(A9,B9) and
A2: j in Seg m;
consider X1 such that
A3: X = X1 and
A4: len X1 = width A9 and
A5: width X1 = width B9 and
A6: A9 * X1 = B9 by A1;
set LAj=Line(A9,j);
set LAi=Line(A9,i);
set RA=RLine(A9,i,LAi+a*LAj);
A7: len (LAi+a*LAj) = width A9 by CARD_1:def 7;
then
A8: len RA = len A9 by MATRIX11:def 3;
set RX=RA*X1;
A9: width RA = width A9 by A7,MATRIX11:def 3;
then
A10: len RX = len RA & width RX=width X1 by A4,MATRIX_3:def 4;
A11: len A9 = len B9 by A1,Th33;
then dom B9=Seg len RA by A8,FINSEQ_1:def 3;
then
A12: Indices RX = Indices B9 by A5,A10,FINSEQ_1:def 3;
set LBj=Line(B9,j);
set LBi=Line(B9,i);
set RB=RLine(B9,i,LBi+a*LBj);
A13: Indices RB = Indices B9 by MATRIX_0:26;
A14: len (LBi+a*LBj) = width B9 by CARD_1:def 7;
then
A15: width RB = width B9 by MATRIX11:def 3;
A16: len (a*LAj) = width A9 & len LAi = width A9 by CARD_1:def 7;
A17: now
A18: rng (a*LBj) c= the carrier of K by FINSEQ_1:def 4;
let o,p be Nat such that
A19: [o,p] in Indices RB;
A20: o in dom B9 by A13,A19,ZFMISC_1:87;
A21: B9*(o,p) = Line(A9,o) "*" Col(X1,p) by A4,A6,A13,A19,MATRIX_3:def 4;
reconsider CX=Col(X1,p) as Element of (width A9)-tuples_on the carrier of
K by A4;
A22: len Col(X1,p)=width A9 by A4,MATRIX_0:def 8;
A23: p in Seg width B9 by A13,A19,ZFMISC_1:87;
then B9*(o,p)=Line(B9,o).p & B9*(j,p)=LBj.p by MATRIX_0:def 7;
then reconsider
LBop = Line(B9,o).p,LBjp = LBj.p as Element of the carrier of K;
p in dom (a*LBj) by A23,FINSEQ_2:124;
then (a*LBj).p in rng (a*LBj) by FUNCT_1:def 3;
then reconsider aLBjp=(a*Line(B9,j)).p as Element of K by A18;
len B9 = m by MATRIX_0:def 2;
then
A24: dom B9=Seg m by FINSEQ_1:def 3;
then [j,p] in Indices B9 by A2,A23,ZFMISC_1:87;
then
A25: B9*(j,p) = LAj"*"Col(X1,p ) by A4,A6,MATRIX_3:def 4;
now
per cases;
suppose
A26: o=i;
then Line(RA,o)=LAi+a*LAj by A7,A20,A24,MATRIX11:28;
hence RX*(o,p) = (LAi+a*LAj) "*" CX by A4,A9,A12,A13,A19,MATRIX_3:def 4
.= Sum(mlt(LAi,CX)+mlt(a*LAj,CX)) by A16,A22,MATRIX_4:56
.= Sum(mlt(LAi,CX)+a*mlt(LAj,CX)) by FVSUM_1:68
.= Sum(mlt(LAi,CX))+Sum(a*mlt(LAj,CX)) by FVSUM_1:76
.= B9*(o,p)+a*(B9*(j,p)) by A21,A25,A26,FVSUM_1:73
.= LBop+a*(B9*(j,p)) by A23,MATRIX_0:def 7
.= LBop+a*LBjp by A23,MATRIX_0:def 7
.= LBop+aLBjp by A23,FVSUM_1:51
.= (LBi+a*LBj).p by A23,A26,FVSUM_1:18
.= RB*(o,p) by A14,A13,A19,A26,MATRIX11:def 3;
end;
suppose
A27: o<>i;
then Line(RA,o)=Line(A9,o) by A20,A24,MATRIX11:28;
hence RX*(o,p) = Line(A9,o)"*"Col(X1,p) by A4,A9,A12,A13,A19,
MATRIX_3:def 4
.= B9*(o,p) by A4,A6,A13,A19,MATRIX_3:def 4
.= RB*(o,p) by A14,A13,A19,A27,MATRIX11:def 3;
end;
end;
hence RB*(o,p) = RX*(o,p);
end;
len RB = len B9 by A14,MATRIX11:def 3;
then RX=RB by A5,A11,A8,A15,A10,A17,MATRIX_0:21;
hence thesis by A3,A4,A5,A9,A15;
end;
Lm5: j in Seg m & i <> j implies Solutions_of(A9,B9) = Solutions_of(RLine(A9,i
,Line(A9,i) + a*Line(A9,j)), RLine(B9,i,Line(B9,i) + a*Line(B9,j)))
proof
assume that
A1: j in Seg m and
A2: i <> j;
set LB=Line(B9,j);
set LA=Line(A9,j);
set RA=RLine(A9,i,Line(A9,i) + a*LA);
set RB=RLine(B9,i,Line(B9,i) + a*LB);
thus Solutions_of(A9,B9) c= Solutions_of(RA,RB)
proof
let x be object such that
A3: x in Solutions_of(A9,B9);
ex X st x=X & len X = width A9 & width X = width B9 & A9 * X = B9 by A3;
hence thesis by A1,A3,Th39;
end;
let x be object such that
A4: x in Solutions_of(RA,RB);
per cases;
suppose
A5: not i in Seg m;
len A9=m by MATRIX_0:def 2;
then len B9=m & RA=A9 by A5,MATRIX13:40,MATRIX_0:def 2;
hence thesis by A4,A5,MATRIX13:40;
end;
suppose
A6: i in Seg m;
reconsider LLA=Line(A9,i)+a*LA,LLB=Line(B9,i)+a*LB, LLRA=Line(RA,i)+(-a)*
Line(RA,j),LLRB=Line(RB,i)+(-a)*Line(RB,j) as Element of (the carrier of K)*
by FINSEQ_1:def 11;
set RRB=RLine(RB,i,Line(RB,i)+(-a)*Line(RB,j));
set RRA=RLine(RA,i,Line(RA,i)+(-a)*Line(RA,j));
A7: ex X st x=X & len X = width RA & width X = width RB & RA * X = RB by A4;
A8: Line(RB,j)=LB by A1,A2,MATRIX11:28;
A9: len (Line(B9,i)+a*LB)=width B9 by CARD_1:def 7;
then
A10: width RB=width B9 by MATRIX11:def 3;
Line(RB,i)=Line(B9,i)+a*LB by A6,A9,MATRIX11:28;
then
A11: Line(RB,i)+(-a)*Line(RB,j)= Line(B9,i)+a*LB+(-1_K*a)*LB by A8,
VECTSP_1:def 8
.= Line(B9,i)+a*LB+((-1_K)*a)*LB by VECTSP_1:9
.= Line(B9,i)+a*LB+(-1_K)*(a*LB) by FVSUM_1:54
.= Line(B9,i)+a*LB+(-(a*LB)) by FVSUM_1:59
.= Line(B9,i)+(a*LB+(-(a*LB))) by FINSEQOP:28
.= Line(B9,i)+width B9|->0.K by FVSUM_1:26
.= Line(B9,i) by FVSUM_1:21;
A12: len (Line(RB,i)+(-a)*Line(RB,j))=width RB by CARD_1:def 7;
then
A13: RRB = Replace(RB,i,LLRB) by MATRIX11:29
.= Replace(Replace(B9,i,LLB),i,LLRB) by A9,MATRIX11:29
.= Replace(B9,i,LLRB) by FUNCT_7:34
.= RLine(B9,i,Line(RB,i)+(-a)*Line(RB,j)) by A12,A10,MATRIX11:29
.= B9 by A11,MATRIX11:30;
A14: Line(RA,j)=LA by A1,A2,MATRIX11:28;
A15: len (Line(A9,i)+a*LA)=width A9 by CARD_1:def 7;
then
A16: width RA=width A9 by MATRIX11:def 3;
Line(RA,i)=Line(A9,i)+a*LA by A6,A15,MATRIX11:28;
then
A17: Line(RA,i)+(-a)*Line(RA,j)= Line(A9,i)+a*LA+(-1_K*a)*LA by A14,
VECTSP_1:def 8
.= Line(A9,i)+a*LA+((-1_K)*a)*LA by VECTSP_1:9
.= Line(A9,i)+a*LA+(-1_K)*(a*LA) by FVSUM_1:54
.= Line(A9,i)+a*LA+(-(a*LA)) by FVSUM_1:59
.= Line(A9,i)+(a*LA+(-(a*LA))) by FINSEQOP:28
.= Line(A9,i)+width A9|->0.K by FVSUM_1:26
.= Line(A9,i) by FVSUM_1:21;
A18: len (Line(RA,i)+(-a)*Line(RA,j))=width RA by CARD_1:def 7;
then RRA = Replace(RA,i,LLRA) by MATRIX11:29
.= Replace(Replace(A9,i,LLA),i,LLRA) by A15,MATRIX11:29
.= Replace(A9,i,LLRA) by FUNCT_7:34
.= RLine(A9,i,Line(RA,i)+(-a)*Line(RA,j)) by A18,A16,MATRIX11:29
.= A9 by A17,MATRIX11:30;
hence thesis by A1,A4,A7,A13,Th39;
end;
end;
theorem Th40:
j in Seg m & (i = j implies a <> -1_K) implies Solutions_of(A9,
B9) = Solutions_of(RLine(A9,i,Line(A9,i) + a*Line(A9,j)), RLine(B9,i,Line(B9,i)
+ a*Line(B9,j)))
proof
assume that
A1: j in Seg m and
A2: i = j implies a <> -1_K;
per cases;
suppose
i<>j;
hence thesis by A1,Lm5;
end;
suppose
A3: i=j;
A4: 1_K+a <> 0.K
proof
assume 1_K+a=0.K;
then -1.K = -1.K+(1_K+a) by RLVECT_1:def 4
.= (-1.K+1_K)+a by RLVECT_1:def 3
.= 0.K+a by VECTSP_1:19
.= a by RLVECT_1:def 4;
hence thesis by A2,A3;
end;
set LB=Line(B9,i);
set LA=Line(A9,i);
A5: LB+a*LB = 1_K*LB+a*LB by FVSUM_1:57
.= (1_K+a)*LB by FVSUM_1:55;
LA+a*LA = 1_K*LA+a*LA by FVSUM_1:57
.= (1_K+a)*LA by FVSUM_1:55;
hence thesis by A3,A4,A5,Lm4;
end;
end;
theorem Th41:
X in Solutions_of(A,B) & i in dom A & Line(A,i) = width A |-> 0.
K implies Line(B,i) = width B |-> 0.K
proof
assume that
A1: X in Solutions_of(A,B) and
A2: i in dom A and
A3: Line(A,i) = width A|->0.K;
set wB0=width B |->0.K;
set LB=Line(B,i);
A4: len LB=width B by CARD_1:def 7;
A5: ex X1 be Matrix of K st X = X1 & len X1 = width A & width X1 = width B &
A * X1 = B by A1;
A6: now
let k such that
A7: 1 <=k & k <= len LB;
A8: k in Seg width B by A4,A7;
len A=len B by A1,Th33;
then dom A=Seg len B by FINSEQ_1:def 3;
then i in dom B by A2,FINSEQ_1:def 3;
then [i,k] in Indices B by A8,ZFMISC_1:87;
then B*(i,k) = Line(A,i) "*" Col(X,k) by A5,MATRIX_3:def 4
.= Sum(0.K*Col(X,k)) by A3,A5,FVSUM_1:66
.= 0.K*Sum(Col(X,k)) by FVSUM_1:73
.= 0.K
.= wB0.k by A8,FINSEQ_2:57;
hence wB0.k = LB.k by A8,MATRIX_0:def 7;
end;
len wB0=width B by CARD_1:def 7;
hence thesis by A4,A6;
end;
Lm6: for nt be Element of n-tuples_on NAT holds i in Seg n implies Line(Segm(A
,nt,Sgm Seg width A),i)=Line(A,nt.i)
proof
let nt be Element of n-tuples_on NAT;
set S=Seg width A;
A1: rng Sgm S=S by FINSEQ_1:def 13;
len Line(A,nt.i)=width A by MATRIX_0:def 7;
then
A2: dom Line(A,nt.i)=S by FINSEQ_1:def 3;
Sgm S = idseq width A by FINSEQ_3:48;
then
A3: Line(A,nt.i) * Sgm S = Line(A,nt.i) by A2,RELAT_1:52;
assume i in Seg n;
hence thesis by A3,A1,MATRIX13:24;
end;
theorem Th42:
for nt be Element of n-tuples_on NAT st rng nt c= dom A & n>0
holds Solutions_of(A,B) c= Solutions_of(Segm(A,nt,Sgm Seg width A), Segm(B,nt,
Sgm Seg width B))
proof
let nt be Element of n-tuples_on NAT such that
A1: rng nt c= dom A and
A2: n>0;
set SA=Segm(A,nt,Sgm Seg width A);
A3: len SA=n by A2,MATRIX_0:23;
width SA=card Seg width A by A2,MATRIX_0:23;
then
A4: width SA=width A by FINSEQ_1:57;
set SB=Segm(B,nt,Sgm Seg width B);
A5: len SB=n by A2,MATRIX_0:23;
width SB=card Seg width B by A2,MATRIX_0:23;
then
A6: width SB=width B by FINSEQ_1:57;
let x be object such that
A7: x in Solutions_of(A,B);
consider X be Matrix of K such that
A8: x = X and
A9: len X = width A and
A10: width X = width B and
A11: A * X = B by A7;
set SX=SA*X;
A12: len A = len B by A7,Th33;
A13: now
A14: len SX=len SA by A9,A4,MATRIX_3:def 4;
let j,k such that
A15: [j,k] in Indices SX;
j in dom SX by A15,ZFMISC_1:87;
then
A16: j in Seg n by A3,A14,FINSEQ_1:def 3;
width SX=width X by A9,A4,MATRIX_3:def 4;
then
A17: k in Seg width B by A10,A15,ZFMISC_1:87;
dom nt=Seg n by FINSEQ_2:124;
then nt.j in rng nt by A16,FUNCT_1:def 3;
then
A18: nt.j in dom A by A1;
dom A=Seg len B by A12,FINSEQ_1:def 3;
then nt.j in dom B by A18,FINSEQ_1:def 3;
then
A19: [nt.j,k] in Indices B by A17,ZFMISC_1:87;
reconsider j9=j,k9=k as Element of NAT by ORDINAL1:def 12;
Sgm Seg width B=idseq width B by FINSEQ_3:48;
then
A20: (Sgm Seg width B).k9=k by A17,FINSEQ_2:49;
j in dom SB by A5,A16,FINSEQ_1:def 3;
then
A21: [j,k] in Indices SB by A6,A17,ZFMISC_1:87;
Line(SA,j)=Line(A,nt.j) by A16,Lm6;
hence SX*(j,k) = Line(A,nt.j) "*" Col(X,k) by A9,A4,A15,MATRIX_3:def 4
.= B*(nt.j9,k) by A9,A11,A19,MATRIX_3:def 4
.= SB*(j,k) by A21,A20,MATRIX13:def 1;
end;
len SX=len SA & width SX=width X by A9,A4,MATRIX_3:def 4;
then SX=SB by A10,A3,A5,A6,A13,MATRIX_0:21;
hence thesis by A8,A9,A10,A4,A6;
end;
theorem Th43:
for nt be Element of n-tuples_on NAT st rng nt c= dom A & dom A
= dom B & n > 0 & for i st i in (dom A) \ rng nt holds Line(A,i) = width A |->
0.K & Line(B,i) = width B |-> 0.K holds Solutions_of(A,B) = Solutions_of(Segm(A
,nt,Sgm Seg width A), Segm(B,nt,Sgm Seg width B))
proof
let nt be Element of n-tuples_on NAT such that
A1: rng nt c= dom A and
A2: dom A=dom B and
A3: n>0 and
A4: for i st i in (dom A) \ rng nt holds Line(A,i) = width A |-> 0.K &
Line(B,i) = width B |-> 0.K;
set SB=Segm(B,nt,Sgm Seg width B);
set SA=Segm(A,nt,Sgm Seg width A);
A5: Solutions_of(SA,SB) c= Solutions_of(A,B)
proof
A6: Seg len A = dom B by A2,FINSEQ_1:def 3;
A7: width SB=card Seg width B by A3,MATRIX_0:23;
then
A8: width SB=width B by FINSEQ_1:57;
let x be object;
assume x in Solutions_of(SA,SB);
then consider X be Matrix of K such that
A9: x = X and
A10: len X = width SA and
A11: width X = width SB and
A12: SA * X = SB;
set AX=A*X;
width SA=card Seg width A by A3,MATRIX_0:23;
then
A13: width SA=width A by FINSEQ_1:57;
then
A14: width AX = width X by A10,MATRIX_3:def 4;
A15: len AX = len A by A10,A13,MATRIX_3:def 4;
A16: now
A17: dom AX=Seg len A by A15,FINSEQ_1:def 3;
let j,k such that
A18: [j,k] in Indices AX;
A19: k in Seg width AX by A18,ZFMISC_1:87;
reconsider j9=j,k9=k as Element of NAT by ORDINAL1:def 12;
A20: j in dom AX by A18,ZFMISC_1:87;
now
per cases;
suppose
A21: j9 in rng nt;
A22: dom nt=Seg n by FINSEQ_2:124;
Sgm Seg width B=idseq width B by FINSEQ_3:48;
then
A23: (Sgm Seg width B).k9=k by A11,A8,A14,A19,FINSEQ_2:49;
consider p be object such that
A24: p in dom nt and
A25: nt.p=j9 by A21,FUNCT_1:def 3;
reconsider p as Element of NAT by A24;
Indices SB =[:Seg n,Seg card Seg width B:] by A3,MATRIX_0:23;
then
A26: [p,k] in Indices SB by A11,A7,A14,A19,A24,A22,ZFMISC_1:87;
Line(SA,p) = Line(A,j9) by A24,A25,A22,Lm6;
hence AX*(j,k) = Line(SA,p)"*"Col(X,k) by A10,A13,A18,MATRIX_3:def 4
.= SB*(p,k9) by A10,A12,A26,MATRIX_3:def 4
.= B*(j,k) by A25,A26,A23,MATRIX13:def 1;
end;
suppose
not j9 in rng nt;
then
A27: j9 in (dom A)\rng nt by A2,A6,A20,A17,XBOOLE_0:def 5;
then
A28: Line(B,j) = width B|->0.K by A4;
Line(A,j) = width A|->0.K by A4,A27;
hence AX*(j,k) = (width A|-> 0.K)"*"Col(X,k) by A10,A13,A18,
MATRIX_3:def 4
.= Sum(0.K*Col(X,k)) by A10,A13,FVSUM_1:66
.= 0.K*Sum(Col(X,k)) by FVSUM_1:73
.= 0.K
.= Line(B,j).k by A11,A8,A14,A19,A28,FINSEQ_2:57
.= B*(j,k) by A11,A8,A14,A19,MATRIX_0:def 7;
end;
end;
hence AX*(j,k)=B*(j,k);
end;
len AX = len B by A15,A6,FINSEQ_1:def 3;
then AX = B by A11,A7,A14,A16,FINSEQ_1:57,MATRIX_0:21;
hence thesis by A9,A10,A11,A13,A8;
end;
Solutions_of(A,B) c= Solutions_of(SA,SB) by A1,A3,Th42;
hence thesis by A5;
end;
theorem Th44:
for N st N c= dom A & N is non empty holds Solutions_of(A,B) c=
Solutions_of(Segm(A,N,Seg width A), Segm(B,N,Seg width B))
proof
let N such that
A1: N c= dom A and
A2: N is non empty;
dom A=Seg len A by FINSEQ_1:def 3;
then rng Sgm N = N by A1,FINSEQ_1:def 13;
hence thesis by A1,A2,Th42;
end;
theorem Th45:
for N st N c= dom A & N is non empty & dom A = dom B & for i st
i in (dom A) \ N holds Line(A,i) = width A |-> 0.K & Line(B,i) = width B |-> 0.
K holds Solutions_of(A,B) = Solutions_of(Segm(A,N,Seg width A), Segm(B,N,Seg
width B))
proof
let N such that
A1: N c= dom A and
A2: N is non empty & dom A=dom B & for i st i in (dom A) \ N holds Line(
A,i) = width A |-> 0.K & Line(B,i) = width B |-> 0.K;
dom A=Seg len A by FINSEQ_1:def 3;
then rng Sgm N = N by A1,FINSEQ_1:def 13;
hence thesis by A1,A2,Th43;
end;
theorem Th46:
i in dom A & len A > 1 implies Solutions_of(A,B) c= Solutions_of
(DelLine(A,i),DelLine(B,i))
proof
assume that
A1: i in dom A and
A2: len A > 1;
reconsider l1=len A-1 as Element of NAT by A2,NAT_1:20;
A3: l1 > 1-1 by A2,XREAL_1:9;
A4: Seg len A=dom A by FINSEQ_1:def 3;
card Seg len A=l1+1 by FINSEQ_1:57;
then card ((Seg len A)\{i})=l1 by A1,A4,STIRL2_1:55;
then
A5: Solutions_of(A,B)c=Solutions_of(Segm(A,(Seg len A)\{i},Seg width A),
Segm(B,(Seg len A)\{i},Seg width B)) by A4,A3,Th44,CARD_1:27,XBOOLE_1:36;
let x be object such that
A6: x in Solutions_of(A,B);
len A=len B by A6,Th33;
then Segm(A,(Seg len A)\{i},Seg width A)=Del(A,i) & Segm(B,(Seg len A)\{i},
Seg width B)=Del(B,i) by MATRIX13:51;
hence thesis by A5,A6;
end;
theorem
for A,B,i st i in dom A & len A > 1 & Line(A,i) = width A |-> 0.K & i
in dom B & Line(B,i) = width B |-> 0.K holds Solutions_of(A,B) = Solutions_of(
DelLine(A,i),DelLine(B,i))
proof
let A,B,i such that
A1: i in dom A and
A2: len A > 1 and
A3: Line(A,i) = width A |-> 0.K and
A4: i in dom B and
A5: Line(B,i) = width B |-> 0.K;
reconsider l1=len A-1 as Element of NAT by A2,NAT_1:20;
A6: l1 > 1-1 by A2,XREAL_1:9;
thus Solutions_of(A,B) c= Solutions_of(DelLine(A,i),DelLine(B,i)) by A1,A2
,Th46;
let x be object such that
A7: x in Solutions_of(DelLine(A,i),DelLine(B,i));
set S=Seg len A;
A8: dom A=S by FINSEQ_1:def 3;
A9: now
let j;
assume j in (dom A)\(S\{i});
then j in dom A /\{i} by A8,XBOOLE_1:48;
then j in {i} by XBOOLE_0:def 4;
hence Line(A,j) = width A |-> 0.K & Line(B,j) = width B |-> 0.K by A3,A5,
TARSKI:def 1;
end;
card S=l1+1 by FINSEQ_1:57;
then
A10: card (S\{i})=l1 by A1,A8,STIRL2_1:55;
(ex mA be Nat st len A = mA + 1 & len Del(A,i) = mA )& ex mB be Nat st
len B = mB + 1 & len Del(B,i) = mB by A1,A4,FINSEQ_3:104;
then
A11: len B=len A by A7,Th33;
then dom A=dom B by A8,FINSEQ_1:def 3;
then
Solutions_of(A,B) = Solutions_of(Segm(A,S\{i},Seg width A),Segm(B,S\{i}
,Seg width B)) by A8,A10,A6,A9,Th45,CARD_1:27,XBOOLE_1:36
.= Solutions_of(DelLine(A,i),Segm(B,S\{i},Seg width B)) by MATRIX13:51
.= Solutions_of(DelLine(A,i),DelLine(B,i)) by A11,MATRIX13:51;
hence thesis by A7;
end;
theorem
for A be Matrix of n,m,K, B be Matrix of n,k,K for P be Function of
Seg n,Seg n holds Solutions_of(A,B) c= Solutions_of(A*P,B*P) & (P is one-to-one
implies Solutions_of(A,B) = Solutions_of(A*P,B*P))
proof
set IDn=idseq n;
len IDn=n & IDn is FinSequence of NAT by CARD_1:def 7,FINSEQ_2:48;
then reconsider IDn as Element of n-tuples_on NAT by FINSEQ_2:92;
let A be Matrix of n,m,K, B be Matrix of n,k,K;
let P be Function of Seg n,Seg n;
A1: rng P c= Seg n by RELAT_1:def 19;
dom IDn=Seg n;
then reconsider IDnP=IDn*P as FinSequence of NAT by FINSEQ_2:47;
dom P=Seg n by FUNCT_2:52;
then n in NAT & dom IDnP = Seg n by A1,ORDINAL1:def 12,RELAT_1:53;
then len IDnP = n by FINSEQ_1:def 3;
then reconsider IDnP as Element of n-tuples_on NAT by FINSEQ_2:92;
A2: n=len A by MATRIX_0:def 2;
A3: (idseq n)*P = P by A1,RELAT_1:53;
then
A4: rng IDnP c=dom A by A1,A2,FINSEQ_1:def 3;
A5: IDn=Sgm Seg n & card Seg n=n by FINSEQ_1:57,FINSEQ_3:48;
then
A6: Segm(A,IDnP,Sgm Seg width A) = Segm(A,Seg len A,Seg width A) * P by A2,
MATRIX13:33
.= A*P by MATRIX13:46;
A7: len B=n by MATRIX_0:def 2;
then
A8: Segm(B,IDnP,Sgm Seg width B) = Segm(B,Seg len B,Seg width B) * P by A5,
MATRIX13:33
.= B*P by MATRIX13:46;
per cases;
suppose
A9: n>0;
hence Solutions_of(A,B)c=Solutions_of(A*P,B*P) by A6,A8,A4,Th42;
A10: card Seg n = card Seg n;
A11: dom A=Seg n by A2,FINSEQ_1:def 3;
A12: dom B=Seg n by A7,FINSEQ_1:def 3;
assume P is one-to-one;
then P is onto by A10,STIRL2_1:60;
then rng P=Seg n by FUNCT_2:def 3;
then for i st i in (dom A) \ rng IDnP holds Line(A,i) = width A |-> 0.K &
Line(B,i) = width B |-> 0.K by A3,A11,XBOOLE_1:37;
hence thesis by A1,A3,A6,A8,A9,A11,A12,Th43;
end;
suppose
A13: n=0;
then len B=0 by MATRIX_0:22;
then
A14: B={};
len A=0 by A13,MATRIX_0:22;
then
A15: A={};
A*P={} by A13;
hence thesis by A15,A14;
end;
end;
theorem Th49:
for A be Matrix of n,m,K, N st card N = n & N c= Seg m & Segm(A,
Seg n,N) = 1.(K,n) & n > 0 ex MVectors be Matrix of m-'n,m,K st Segm(MVectors,
Seg (m-'n),Seg m \ N) = 1.(K,m-'n) & Segm(MVectors,Seg (m-'n),N) = -Segm(A,Seg
n,Seg m \N)@ & for l for M be Matrix of m,l,K st for i st i in Seg l holds (ex
j st j in Seg (m-'n) & Col(M,i) = Line(MVectors,j)) or Col(M,i) = m|->0.K holds
M in Solutions_of(A,0.(K,n,l))
proof
let A be Matrix of n,m,K,N such that
A1: card N = n and
A2: N c= Seg m and
A3: Segm(A,Seg n,N) = 1.(K,n) and
A4: n>0;
set SN=Seg m\N;
A5: m-'n=0 or m-'n > 0;
A6: card Seg m = m by FINSEQ_1:57;
then
A7: card SN=m-n by A1,A2,CARD_2:44;
set ZERO=0.(K,m-'n,m);
A8: SN c= Seg m by XBOOLE_1:36;
A9: now
per cases;
suppose
m-'n=0;
then Seg (m-'n) = {};
then [:Seg (m-'n),N:] = {} & [:Seg (m-'n),SN:] ={} by ZFMISC_1:90;
hence
[:Seg (m-'n),N:]c=Indices ZERO & [:Seg (m-'n),SN:]c=Indices ZERO;
end;
suppose
m-'n>0;
then Indices ZERO = [:Seg (m-'n),Seg m:] by MATRIX_0:23;
hence
[:Seg (m-'n),N:]c=Indices ZERO & [:Seg (m-'n),SN:]c=Indices ZERO by A2,A8
,ZFMISC_1:96;
end;
end;
set SA=Segm(A,Seg n,SN);
card Seg n = n by FINSEQ_1:57;
then
A10: len SA=n by A4,MATRIX_0:23;
A11: len (SA@)=len (-SA@) & width (SA@)=width (-SA@) by MATRIX_3:def 2;
A12: width A=m by A4,MATRIX_0:23;
Segm n c= Segm card Seg m by A1,A2,CARD_1:11;
then
A13: n <= m by A6,NAT_1:39;
then
A14: m-'n=m-n by XREAL_1:233;
then width SA=m-'n by A4,A7,MATRIX_0:23;
then len (SA@)=0 & m-'n=0 or len (SA@)=m-'n & width (SA@)=n by A10,A5,
MATRIX_0:54,def 6;
then -SA@={} &m-'n=0 or -SA@ is Matrix of m-'n,n,K by A11,MATRIX_0:51;
then reconsider SAT=-SA@ as Matrix of m-'n,n,K by MATRIX_0:13;
set ONE= 1.(K,m-'n);
A15: N misses SN by XBOOLE_1:79;
[:Seg (m-'n),N:]/\[:Seg (m-'n),SN:] = [:Seg (m-'n),N/\SN:] by ZFMISC_1:99
.= [:Seg (m-'n),{}:] by A15
.= {} by ZFMISC_1:90;
then
card Seg (m-'n) = m-'n & for i,j,bi,bj,ci,cj be Nat st [i,j] in [:Seg (
m-'n) ,N:]/\[:Seg (m-'n),SN:] & bi = Sgm (Seg (m-'n))".i & bj = Sgm N".j & ci =
Sgm ( Seg (m-'n))".i & cj = Sgm SN".j holds SAT*(bi,bj) = ONE*(ci,cj) by
FINSEQ_1:57;
then consider V be Matrix of len ZERO,width ZERO,K such that
A16: Segm(V,Seg(m-'n),N) = SAT and
A17: Segm(V,Seg (m-'n),SN) = ONE and
for i,j st [i,j] in Indices V\([:Seg (m-'n),N:]\/[:Seg (m-'n),SN:])
holds V*(i,j) = ZERO*(i,j) by A1,A9,A14,A7,Th9;
m-'n=0 or m-'n > 0;
then
len ZERO=0 & m-'n=0 & len V=len ZERO or len ZERO=m-'n & width ZERO=m by
MATRIX_0:23,def 2;
then V={} & m-'n=0 or V is Matrix of m-'n,m,K;
then reconsider V as Matrix of m-'n,m,K;
take V;
thus Segm(V,Seg (m-'n),SN) = ONE & Segm(V,Seg (m-'n),N)=-SA@ by A16,A17;
let l;
let M be Matrix of m,l,K such that
A18: for i st i in Seg l holds (ex j st j in Seg (m-'n) & Col(M,i) =
Line(V,j)) or Col(M,i) = m|->0.K;
set Z=0.(K,n,l);
A19: len M=m by A4,A13,MATRIX_0:23;
A20: width M=l by A4,A13,MATRIX_0:23;
then
A21: width (A*M)=l by A12,A19,MATRIX_3:def 4;
len A=n by A4,MATRIX_0:23;
then len (A*M)=n by A12,A19,MATRIX_3:def 4;
then reconsider AM=A*M as Matrix of n,l,K by A21,MATRIX_0:51;
A22: Indices A=[:Seg n,Seg m:] by A4,MATRIX_0:23;
then
A23: [:Seg n,N:] c= Indices A by A2,ZFMISC_1:96;
now
A24: Indices AM=Indices Z by MATRIX_0:26;
let i,j such that
A25: [i,j] in Indices AM;
reconsider I=i,J=j as Element of NAT by ORDINAL1:def 12;
A26: Indices AM=[:Seg n,Seg l:] by A4,MATRIX_0:23;
then
A27: I in Seg n by A25,ZFMISC_1:87;
A28: J in Seg l by A25,A26,ZFMISC_1:87;
now
per cases by A18,A28;
suppose
ex jj be Nat st jj in Seg (m-'n) & Col(M,J) = Line(V,jj);
then consider jj be Nat such that
A29: jj in Seg (m-'n) and
A30: Col(M,J) = Line(V,jj);
A31: jj = idseq (m-'n).jj by A29,FINSEQ_2:49
.= Sgm (Seg (m-'n)).jj by FINSEQ_3:48;
A32: Indices ONE=[:Seg (m-'n),Seg (m-'n):] by MATRIX_0:24;
then
A33: [jj,jj] in Indices ONE by A29,ZFMISC_1:87;
A34: rng Sgm SN=SN by A8,FINSEQ_1:def 13;
A35: dom Sgm SN=Seg(m-'n) by A14,A7,FINSEQ_3:40,XBOOLE_1:36;
then
A36: Sgm SN.jj in SN by A29,A34,FUNCT_1:def 3;
then
A37: Line(A,I).(Sgm SN.jj)=A*(I,Sgm SN.jj) by A12,A8,MATRIX_0:def 7;
A38: m-'n<>0 by A29;
then
A39: width V=m by MATRIX_0:23;
then
A40: Line(V,jj).(Sgm SN.jj)=V*(jj,Sgm SN.jj) by A8,A36,MATRIX_0:def 7
.= ONE*(jj,jj) by A17,A31,A33,MATRIX13:def 1
.= 1_K by A33,MATRIX_1:def 3;
A41: len Line(A,I)=m by A12,MATRIX_0:def 7;
A42: I = idseq n.I by A27,FINSEQ_2:49
.= Sgm (Seg n).I by FINSEQ_3:48;
len Line(V,jj)=m by A39,MATRIX_0:def 7;
then len mlt(Line(A,I),Line(V,jj))=m by A41,MATRIX_3:6;
then
A43: dom mlt(Line(A,I),Line(V,jj))=Seg m by FINSEQ_1:def 3;
then
A44: mlt(Line(A,I),Line(V,jj))/.(Sgm SN.jj) = mlt(Line(A,I),Line(V,jj)
).(Sgm SN.jj) by A8,A36,PARTFUN1:def 6
.= (A*(I,Sgm SN.jj))*1_K by A12,A8,A39,A36,A37,A40,FVSUM_1:61
.=A*(I,Sgm SN.jj) by VECTSP_1:def 6;
A45: Indices V=Indices ZERO & rng Sgm Seg(m-'n)=Seg(m-'n) by FINSEQ_1:def 13
,MATRIX_0:26;
A46: rng Sgm N=N by A2,FINSEQ_1:def 13;
A47: rng Sgm Seg n=Seg n by FINSEQ_1:def 13;
A48: now
let kk be Nat such that
A49: kk in Seg m and
A50: kk<> Sgm SN.jj and
A51: kk<>Sgm N.I;
now
per cases by A49,XBOOLE_0:def 5;
suppose
kk in N;
then consider x being object such that
A52: x in dom Sgm N and
A53: Sgm N.x=kk by A46,FUNCT_1:def 3;
reconsider x as Element of NAT by A52;
A54: Line(V,jj).(Sgm N.x) = V*(jj,Sgm N.x) by A39,A49,A53,
MATRIX_0:def 7;
[Sgm (Seg n).I,Sgm N.x] in Indices A by A22,A27,A42,A49,A53,
ZFMISC_1:87;
then
A55: [I,x] in Indices Segm(A,Seg n,N) by A23,A46,A47,MATRIX13:17;
Line(A,I).(Sgm N.x) = A*(I,Sgm N.x) by A12,A49,A53,MATRIX_0:def 7
.= Segm(A,Seg n,N)*(I,x) by A42,A55,MATRIX13:def 1
.= 0.K by A3,A51,A53,A55,MATRIX_1:def 3;
hence mlt(Line(A,I),Line(V,jj)).kk = 0.K*(V*(jj,Sgm N.x)) by A12
,A39,A49,A53,A54,FVSUM_1:61
.= 0.K;
end;
suppose
kk in SN;
then consider x being object such that
A56: x in dom Sgm SN and
A57: Sgm SN.x=kk by A34,FUNCT_1:def 3;
reconsider x as Element of NAT by A56;
A58: Line(A,I).(Sgm SN.x)=A*(I,Sgm SN.x) by A12,A49,A57,MATRIX_0:def 7
;
A59: [jj,x] in Indices ONE by A29,A35,A32,A56,ZFMISC_1:87;
Line(V,jj).(Sgm SN.x) = V*(jj,Sgm SN.x) by A39,A49,A57,
MATRIX_0:def 7
.= ONE*(jj,x) by A17,A31,A59,MATRIX13:def 1
.= 0.K by A50,A57,A59,MATRIX_1:def 3;
hence mlt(Line(A,I),Line(V,jj)).kk = (A*(I,Sgm SN.x))*0.K by A12
,A39,A49,A57,A58,FVSUM_1:61
.= 0.K;
end;
end;
hence mlt(Line(A,I),Line(V,jj)).kk=0.K;
end;
dom Sgm N=Seg n by A1,A2,FINSEQ_3:40;
then
A60: Sgm N.I in N by A27,A46,FUNCT_1:def 3;
then
A61: Sgm SN.jj <> Sgm N.I by A15,A36,XBOOLE_0:3;
[Sgm (Seg n).I,Sgm N.I] in Indices A by A2,A22,A27,A60,A42,ZFMISC_1:87;
then
A62: [I,I] in Indices Segm(A,Seg n,N) by A23,A46,A47,MATRIX13:17;
Indices V=[:Seg (m-'n),Seg m:] by A38,MATRIX_0:23;
then [Sgm (Seg (m-'n)).jj,Sgm N.I] in Indices V by A2,A29,A60,A31,
ZFMISC_1:87;
then
A63: [jj,I] in Indices Segm(V,Seg(m-'n),N) by A9,A46,A45,MATRIX13:17;
A64: Indices SAT= Indices (SA@) by Lm1;
then
A65: [I,jj] in Indices SA by A16,A63,MATRIX_0:def 6;
A66: Line(V,jj).(Sgm N.I) = V*(jj,Sgm N.I) by A2,A39,A60,MATRIX_0:def 7
.= Segm(V,Seg(m-'n),N)*(jj,I) by A31,A63,MATRIX13:def 1
.= -((SA@)*(jj,I)) by A16,A63,A64,MATRIX_3:def 2
.= -(SA*(I,jj)) by A65,MATRIX_0:def 6
.= -(A*(I,Sgm SN.jj)) by A42,A65,MATRIX13:def 1;
A67: Line(A,I).(Sgm N.I) = A*(I,Sgm N.I) by A2,A12,A60,MATRIX_0:def 7
.= Segm(A,Seg n,N)*(I,I) by A42,A62,MATRIX13:def 1
.= 1_K by A3,A62,MATRIX_1:def 3;
mlt(Line(A,I),Line(V,jj))/.(Sgm N.I) = mlt(Line(A,I),Line(V,jj)).
(Sgm N.I) by A2,A43,A60,PARTFUN1:def 6
.= 1_K*(-(A*(I,Sgm SN.jj))) by A2,A12,A39,A60,A67,A66,FVSUM_1:61
.= -(A*(I,Sgm SN.jj)) by VECTSP_1:def 6;
then Sum(mlt(Line(A,I),Line(V,jj))) = A*(I,Sgm SN.jj)+(-(A*(I,Sgm SN.
jj))) by A2,A8,A43,A60,A36,A44,A61,A48,Th7
.= 0.K by VECTSP_1:16;
hence Z*(i,j) = Line(A,I)"*"Line(V,jj) by A25,A24,MATRIX_3:1
.= AM*(i,j) by A12,A19,A25,A30,MATRIX_3:def 4;
end;
suppose
Col(M,J) = m|->0.K;
hence AM*(i,j) = Line(A,I)"*"(m|->0.K) by A12,A19,A25,MATRIX_3:def 4
.= Sum(0.K * Line(A,I)) by A12,FVSUM_1:66
.= 0.K*Sum(Line(A,I)) by FVSUM_1:73
.= 0.K
.= Z*(i,j) by A25,A24,MATRIX_3:1;
end;
end;
hence AM*(i,j) = Z*(i,j);
end;
then AM = Z by MATRIX_0:27;
hence thesis by A12,A19,A20,A21;
end;
theorem Th50:
for A be Matrix of n,m,K, B be Matrix of n,l,K, N st card N = n
& N c= Seg m & n>0 & Segm(A,Seg n,N) = 1.(K,n) ex X be Matrix of m,l,K st Segm(
X,Seg m \ N,Seg l) = 0.(K,m-'n,l) & Segm(X,N,Seg l) = B & X in Solutions_of(A,B
)
proof
let A be Matrix of n,m,K, B be Matrix of n,l,K, N such that
A1: card N = n and
A2: N c= Seg m and
A3: n>0 and
A4: Segm(A,Seg n,N) = 1.(K,n);
A5: width A=m by A3,MATRIX_0:23;
set Z=0.(K,m,l);
set SN=Seg m\N;
A6: card Seg m=m by FINSEQ_1:57;
then
A7: m-'n=m-n & card SN=m-n by A1,A2,CARD_2:44,NAT_1:43,XREAL_1:233;
set ZERO=0.(K,m-'n,l);
A8: N misses SN by XBOOLE_1:79;
[:N,Seg l:]/\[:SN,Seg l:] = [:N/\SN,Seg l:] by ZFMISC_1:99
.= [:{},Seg l:] by A8
.= {} by ZFMISC_1:90;
then
A9: for i,j,bi,bj,ci,cj be Nat st [i,j] in [:N,Seg l:]/\[:SN,Seg l:] & bi =
Sgm N".i & bj = Sgm (Seg l)".j & ci = Sgm SN".i & cj = Sgm (Seg l)".j holds B*(
bi,bj) = ZERO*(ci,cj);
A10: Indices A=[:Seg n,Seg m:] by A3,MATRIX_0:23;
A11: n <=card Seg m by A1,A2,NAT_1:43;
then
A12: len Z=m & width Z=l by A3,A6,MATRIX_0:23;
A13: Indices Z=[:Seg m,Seg l:] by A3,A11,A6,MATRIX_0:23;
then
A14: [:N,Seg l:] c= Indices Z by A2,ZFMISC_1:95;
A15: SN c= Seg m by XBOOLE_1:36;
then card Seg l=l & [:SN,Seg l:] c= Indices Z by A13,FINSEQ_1:57,ZFMISC_1:95;
then consider X be Matrix of m,l,K such that
A16: Segm(X,N,Seg l) = B and
A17: Segm(X,SN,Seg l) = ZERO and
for i,j st [i,j] in Indices X\([:N,Seg l:]\/[:SN,Seg l:]) holds X*(i,j)
= Z*(i,j) by A1,A7,A12,A14,A9,Th9;
set AX=A*X;
A18: len X=m by A3,A11,A6,MATRIX_0:23;
then
A19: dom X=Seg m by FINSEQ_1:def 3;
len A=n by A3,MATRIX_0:23;
then
A20: len AX=n by A5,A18,MATRIX_3:def 4;
take X;
thus Segm(X,SN,Seg l) = ZERO & Segm(X,N,Seg l) = B by A16,A17;
A21: Indices X=[:Seg m,Seg l:] by A3,A11,A6,MATRIX_0:23;
A22: width B=l by A3,MATRIX_0:23;
A23: width X=l by A3,A11,A6,MATRIX_0:23;
then width AX=l by A5,A18,MATRIX_3:def 4;
then reconsider AX as Matrix of n,l,K by A20,MATRIX_0:51;
A24: Indices B=[:Seg n,Seg l:] by A3,MATRIX_0:23;
now
A25: [:N,Seg l:] c= Indices X by A2,A21,ZFMISC_1:95;
let i,j such that
A26: [i,j] in Indices AX;
reconsider I=i,J=j as Element of NAT by ORDINAL1:def 12;
A27: Indices AX=Indices B by MATRIX_0:26;
then
A28: i in Seg n by A24,A26,ZFMISC_1:87;
len Line(A,i)=m & len Col(X,j)=m by A5,A18,CARD_1:def 7;
then len mlt(Line(A,i),Col(X,j))=m by MATRIX_3:6;
then
A29: dom mlt(Line(A,i),Col(X,j))=Seg m by FINSEQ_1:def 3;
A30: rng Sgm (Seg l)=Seg l by FINSEQ_1:def 13;
A31: rng Sgm (Seg n)=Seg n & [:Seg n,N:] c= Indices A by A2,A10,FINSEQ_1:def 13
,ZFMISC_1:95;
A32: rng Sgm N=N by A2,FINSEQ_1:def 13;
dom Sgm N=Seg n by A1,A2,FINSEQ_3:40;
then
A33: Sgm N.i in N by A28,A32,FUNCT_1:def 3;
A34: j in Seg l by A22,A26,A27,ZFMISC_1:87;
then
A35: j = (idseq l).j by FINSEQ_2:49
.= Sgm (Seg l).j by FINSEQ_3:48;
then [Sgm N.I,Sgm(Seg l).J] in Indices X by A2,A21,A34,A33,ZFMISC_1:87;
then
A36: [I,J] in Indices B by A16,A32,A30,A25,MATRIX13:17;
A37: rng Sgm SN=SN by A15,FINSEQ_1:def 13;
A38: i = (idseq n).i by A28,FINSEQ_2:49
.= Sgm (Seg n).i by FINSEQ_3:48;
then [Sgm (Seg n).i,Sgm N.i] in Indices A by A2,A10,A28,A33,ZFMISC_1:87;
then
A39: [I,I] in Indices 1.(K,n) by A4,A32,A31,MATRIX13:17;
A40: [:SN,Seg l:] c= Indices X by A15,A21,ZFMISC_1:95;
A41: now
let kk be Nat such that
A42: kk in Seg m and
A43: kk <> Sgm N.I;
per cases by A42,XBOOLE_0:def 5;
suppose
A44: kk in N;
then consider x being object such that
A45: x in dom Sgm N and
A46: Sgm N.x=kk by A32,FUNCT_1:def 3;
reconsider x as Element of NAT by A45;
[Sgm (Seg n).i,Sgm N.x] in Indices A by A2,A10,A28,A38,A44,A46,
ZFMISC_1:87;
then
A47: [I,x] in Indices 1.(K,n) by A4,A32,A31,MATRIX13:17;
A48: Col(X,j).kk = X*(kk,j) by A2,A19,A44,MATRIX_0:def 8;
Line(A,i).(Sgm N.x) = A*(I,Sgm N.x) by A2,A5,A44,A46,MATRIX_0:def 7
.= (Segm(A,Seg n,N))*(I,x) by A4,A38,A47,MATRIX13:def 1
.= 0.K by A4,A43,A46,A47,MATRIX_1:def 3;
hence
mlt(Line(A,i),Col(X,j)).kk = 0.K*(X*(kk,j)) by A2,A5,A18,A44,A46,A48,
FVSUM_1:61
.= 0.K;
end;
suppose
A49: kk in SN;
then consider x being object such that
A50: x in dom Sgm SN and
A51: Sgm SN.x=kk by A37,FUNCT_1:def 3;
reconsider x as Element of NAT by A50;
A52: Line(A,i).kk=A*(I,Sgm SN.x) by A5,A42,A51,MATRIX_0:def 7;
[Sgm SN.x,Sgm (Seg l).J] in Indices X by A15,A21,A34,A35,A49,A51,
ZFMISC_1:87;
then
A53: [x,J] in Indices ZERO by A17,A30,A37,A40,MATRIX13:17;
Col(X,j).kk = X*(Sgm SN.x,Sgm (Seg l).j) by A15,A19,A35,A49,A51,
MATRIX_0:def 8
.= ZERO*(x,J) by A17,A53,MATRIX13:def 1
.= 0.K by A53,MATRIX_3:1;
hence
mlt(Line(A,i),Col(X,j)).kk = (A*(I,Sgm SN.x))*0.K by A5,A18,A42,A52,
FVSUM_1:61
.= 0.K;
end;
end;
A54: Col(X,j).(Sgm N.i) = X*(Sgm N.i,j) by A2,A19,A33,MATRIX_0:def 8
.= B*(I,J) by A16,A35,A36,MATRIX13:def 1;
Line(A,i).(Sgm N.i) = A*(I,Sgm N.I) by A2,A5,A33,MATRIX_0:def 7
.= (Segm(A,Seg n,N))*(I,I) by A4,A38,A39,MATRIX13:def 1
.= 1_K by A4,A39,MATRIX_1:def 3;
then
A55: mlt(Line(A,i),Col(X,j)).(Sgm N.i) = 1_K * (B*(I,J)) by A2,A5,A18,A33,A54,
FVSUM_1:61
.= B*(I,J) by VECTSP_1:def 4;
AX*(i,j) = Line(A,i)"*"Col(X,j) by A5,A18,A26,MATRIX_3:def 4
.= Sum(mlt(Line(A,i),Col(X,j)));
hence AX*(i,j)=B*(i,j) by A2,A33,A55,A29,A41,MATRIX_3:12;
end;
then AX=B by MATRIX_0:27;
hence thesis by A5,A22,A18,A23;
end;
theorem Th51:
for A be Matrix of 0,n,K, B be Matrix of 0,m,K holds Solutions_of(A,B) = {{}}
proof
let A be Matrix of 0,n,K, B be Matrix of 0,m,K;
A1: len A=0 by MATRIX_0:def 2;
A2: Solutions_of(A,B) c= {{}}
proof
let x be object;
assume x in Solutions_of(A,B);
then ex X st X=x &len X = width A & width X = width B & A * X = B;
then x={} by A1,MATRIX_0:def 3;
hence thesis by TARSKI:def 1;
end;
len B=0 by MATRIX_0:def 2;
then
A3: B={} & width B=0 by MATRIX_0:def 3;
A4: width A=0 by A1,MATRIX_0:def 3;
then len (A*A)=0 by A1,MATRIX_3:def 4;
then A*A={};
then A in Solutions_of(A,B) by A1,A4,A3;
hence thesis by A2,ZFMISC_1:33;
end;
theorem Th52:
for B be Matrix of K st Solutions_of(0.(K,n,k),B) is non empty
holds B = 0.(K,n,width B)
proof
let B be Matrix of K;
set A=0.(K,n,k);
set ZERO=0.(K,n,width B);
assume Solutions_of(0.(K,n,k),B) is non empty;
then consider x being object such that
A1: x in Solutions_of(0.(K,n,k),B);
A2: len A=n by MATRIX_0:def 2;
then
A3: dom A=Seg n by FINSEQ_1:def 3;
A4: len ZERO=n by MATRIX_0:def 2;
then
A5: len B=len ZERO by A1,A2,Th33;
then reconsider B9=B as Matrix of n,width B,K by A4,MATRIX_0:51;
A6: ex X st X=x & len X = width A & width X = width B & A * X = B by A1;
now
let i such that
A7: 1<=i & i<=n;
A8: width A=k by A7,MATRIX_0:23;
A9: i in Seg n by A7;
then Line(A,i) = A.i by MATRIX_0:52
.= width A|->0.K by A9,A8,FINSEQ_2:57;
then width B|-> 0.K = Line(B,i) by A1,A6,A3,A9,Th41
.= B9.i by A9,MATRIX_0:52;
hence B.i=ZERO.i by A9,FINSEQ_2:57;
end;
hence thesis by A4,A5;
end;
theorem Th53:
for A be Matrix of n,k,K, B be Matrix of n,m,K st n > 0 holds x
in Solutions_of(A,B) implies x is Matrix of k,m,K
proof
let A be Matrix of n,k,K, B be Matrix of n,m,K;
assume n > 0;
then
A1: width A=k & width B=m by MATRIX_0:23;
assume x in Solutions_of(A,B);
then ex X st X=x & len X = k & width X = m & A * X = B by A1;
hence thesis by MATRIX_0:51;
end;
theorem Th54:
n > 0 & k > 0 implies Solutions_of(0.(K,n,k),0.(K,n,m)) = the set of all X
where X is Matrix of k,m,K
proof
assume that
A1: n > 0 and
A2: k > 0;
set B=0.(K,n,m);
A3: width B=m by A1,MATRIX_0:23;
set XX=the set of all X where X is Matrix of k,m,K;
set A=0.(K,n,k);
thus Solutions_of(A,B) c= XX
proof
let x be object;
assume x in Solutions_of(A,B);
then x is Matrix of k,m,K by A1,Th53;
hence thesis;
end;
let x be object;
assume x in XX;
then consider X be Matrix of k,m,K such that
A4: x=X and
not contradiction;
A5: width A=k & len X=k by A1,A2,MATRIX_0:23;
A6: width X=m by A2,MATRIX_0:23;
len A=n by A1,MATRIX_0:23;
then A*X=B by A1,A2,A5,A6,MATRIX_5:22;
hence thesis by A4,A3,A5,A6;
end;
theorem
n>0 & Solutions_of(0.(K,n,0),0.(K,n,m)) is non empty implies m = 0
proof
assume that
A1: n>0 and
A2: Solutions_of(0.(K,n,0),0.(K,n,m)) is non empty;
consider x being object such that
A3: x in Solutions_of(0.(K,n,0),0.(K,n,m)) by A2;
A4: width 0.(K,n,0)=0 by A1,MATRIX_0:23;
ex X st X=x & len X = width 0.(K,n,0) & width X = width 0.(K,n,m) & 0.(K,
n,0) * X = 0.(K,n,m) by A3;
hence 0 = width 0.(K,n,m) by A4,MATRIX_0:def 3
.= m by A1,MATRIX_0:23;
end;
theorem Th56:
Solutions_of(0.(K,n,0),0.(K,n,0)) = {{}}
proof
per cases;
suppose
n=0;
hence thesis by Th51;
end;
suppose
A1: n>0;
set B=0.(K,n,0);
set A=0.(K,n,0);
reconsider E={} as Matrix of 0,0,K by MATRIX_0:13;
A2: width A=0 by A1,MATRIX_0:23;
then
A3: for i,j st [i,j] in Indices B holds B*(i,j)=(A*E)*(i,j) by ZFMISC_1:90;
A4: Solutions_of(A,B) c= {{}}
proof
let x be object;
assume x in Solutions_of(A,B);
then reconsider X=x as Matrix of 0,0,K by A1,Th53;
len X=0 by MATRIX_0:def 2;
then X={};
hence thesis by TARSKI:def 1;
end;
A5: len E=0;
A6: width E=0 by MATRIX_0:24;
then
A7: width (A*E)=0 by A2,A5,MATRIX_3:def 4;
A8: len A=n by A1,MATRIX_0:23;
then len (A*E)=n by A2,A5,MATRIX_3:def 4;
then A*E=B by A8,A2,A7,A3,MATRIX_0:21;
then E in Solutions_of(A,B) by A2,A5,A6;
hence thesis by A4,ZFMISC_1:33;
end;
end;
begin ::Gaussian eliminations
scheme
GAUSS1{ K() -> Field, n,m,m9() -> Nat, A() -> Matrix of n(),m(),K(), B() ->
Matrix of n(),m9(),K(), F(Matrix of n(),m9(),K(),Nat,Nat,Element of K())->
Matrix of n(),m9(),K(), P[set,set]}: ex A9 be Matrix of n(),m(),K(), B9 be
Matrix of n(),m9(),K(), N be without_zero finite Subset of NAT st N c= Seg m()
& the_rank_of A() = the_rank_of A9 & the_rank_of A() = card N & P[A9,B9] & Segm
(A9,Seg card N,N) is diagonal & ( for i st i in Seg card N holds A9*(i,Sgm N/.i
) <> 0.K() ) & ( for i st i in dom A9 & i > card N holds Line(A9,i)= m() |->
0.K(
) ) & for i,j st i in Seg card N & j in Seg width A9 & j < Sgm N.i holds A9*(i,
j) = 0.K()
provided
A1: P[A(),B()] and
A2: for A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() st P[
A9,B9] for i,j st i <> j & j in dom A9 for a be Element of K() holds P[RLine(A9
,i,Line(A9,i) + a*Line(A9,j)),F(B9,i,j,a)]
proof
defpred PP[FinSequence of NAT,Nat,Nat, Matrix of n(),m(),K()] means $4*($2,
$1/.$2) <> 0.K() & ( $3 in dom $1 & $2 < $3 implies $1/.$2 < $1/.$3 )& ( $3 in
dom $1 \ {$2} implies $4*($3,$1/.$2) = 0.K() )& ( $3 in Seg width $4 & $3 < $1
/.$2 implies $4*($2,$3) = 0.K());
set r = the_rank_of A();
ex A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() st P[A9,B9]
& r = the_rank_of A9 & (for i st i in dom A9 & i > r holds Line(A9,i)=m() |->
0.K
())& ex f be FinSequence of NAT st len f = the_rank_of A9 & f is one-to-one &
rng f c= Seg width A9 & for i,j st i in dom f holds PP[f,i,j,A9]
proof
per cases;
suppose
A3: n()=0;
take A9=A(),B9=B();
dom A9=Seg len A9 & len A9=0 by A3,FINSEQ_1:def 3,MATRIX_0:def 2;
hence P[A9,B9] & r = the_rank_of A9 & for i st i in dom A9 & i > r holds
Line(A9,i)=m() |->0.K() by A1;
take <*>NAT;
len A9=0 by A3,MATRIX_0:22;
hence thesis by MATRIX13:74;
end;
suppose
A4: n()>0;
defpred Q[Nat] means $1 <= m() implies ex A9 be Matrix of n(),m(),K(),
B9 be Matrix of n(),m9(),K() st P[A9,B9] & r = the_rank_of A9 & ex f be
FinSequence of NAT st (for i,j st [i,j] in Indices A9&i>len f&j<=$1 holds A9*(i
,j)=0.K())& f is one-to-one & len f <= $1 & len f<=n() & rng f c= Seg $1 & for
i,j st i in dom f holds PP[f,i,j,A9];
A5: for n st Q[n] holds Q[n+1]
proof
let n such that
A6: Q[n];
set n1=n+1;
assume
A7: n1<=m();
then consider
A9 be Matrix of n(),m(),K(),B9 be Matrix of n(),m9(),K() such
that
A8: ( P[A9,B9])& r = the_rank_of A9 and
A9: ex f be FinSequence of NAT st (for i,j st [i,j] in Indices A9
&i>len f&j<=n holds A9*(i,j)=0.K())& f is one-to-one & len f <= n &len f<=n()&
rng f c= Seg n & for i,j st i in dom f holds PP[f,i,j,A9] by A6,NAT_1:13;
consider f be FinSequence of NAT such that
A10: for i,j st [i,j] in Indices A9&i>len f&j<=n holds A9*(i,j)=0. K() and
A11: f is one-to-one and
A12: len f <= n and
A13: len f<=n() and
A14: rng f c=Seg n and
A15: for i,j st i in dom f holds PP[f,i,j,A9] by A9;
per cases;
suppose
A16: for i,j st [i,j] in Indices A9 & i>len f & j=n1 holds A9*(i
,j)=0.K();
A17: now
let i,j such that
A18: [i,j] in Indices A9 & i>len f and
A19: j<=n1;
j<=n or j=n1 by A19,NAT_1:8;
hence A9*(i,j)=0.K() by A10,A16,A18;
end;
n<=n1 by NAT_1:13;
then Seg n c= Seg n1 by FINSEQ_1:5;
then
A20: rng f c= Seg n1 by A14;
len f <=n1 by A12,NAT_1:12;
hence thesis by A8,A11,A13,A15,A17,A20;
end;
suppose
ex i,j st [i,j] in Indices A9 &i>len f&j=n1&A9*(i,j)<>0.K();
then consider i0,j0 be Nat such that
A21: [i0,j0] in Indices A9 and
A22: i0>len f and
A23: j0=n1 and
A24: A9*(i0,j0)<>0.K();
A25: Indices A9=[:Seg n(),Seg m():] by A4,MATRIX_0:23;
then
A26: n1 in Seg m() by A21,A23,ZFMISC_1:87;
A27: i0 in Seg n() by A21,A25,ZFMISC_1:87;
then
A28: i0<=n() by FINSEQ_1:1;
A29: len f+1<=i0 by A22,NAT_1:13;
then 0+1<=len f+1 & len f+1<=n() by A28,XREAL_1:7,XXREAL_0:2;
then
A30: len f+1 in Seg n();
defpred QQ[Nat] means $1<=n() implies ex A9 be Matrix of n(),m(),K()
, B9 be Matrix of n(),m9(),K() st P[A9,B9] & r = the_rank_of A9 & A9*(len f+1,
n1) <> 0.K() & (for i,j st [i,j] in Indices A9 & i>len f&j<=n holds A9*(i,j)=0.
K())& (for i,j st i in dom f holds PP[f,i,j,A9]) & for j st j in dom A9\{len f+
1} & j<=$1 holds A9*(j,n1)=0.K();
A31: dom f = Seg len f by FINSEQ_1:def 3;
n<=m() by A7,NAT_1:13;
then
A32: Seg n c= Seg m() by FINSEQ_1:5;
A33: Seg len f c=Seg n() by A13,FINSEQ_1:5;
A34: for k st QQ[k] holds QQ[k+1]
proof
let k such that
A35: QQ[k];
set k1=k+1;
assume k1<=n();
then consider
AA be Matrix of n(),m(),K(), BB be Matrix of n(),m9( ),K(
) such that
A36: P[AA,BB] and
A37: r = the_rank_of AA and
A38: AA*(len f+1,n1) <> 0.K() and
A39: for i,j st [i,j] in Indices AA & i>len f&j<=n holds AA*(
i,j)=0.K() and
A40: for i,j st i in dom f holds PP[f,i,j,AA] and
A41: for j st j in dom AA\{len f+1} & j<=k holds AA*(j,n1)=0.
K() by A35,NAT_1:13;
now
per cases;
suppose
A42: k1=len f+1;
take RA=AA,RB=BB;
now
let j such that
A43: j in dom RA \ {len f+1} and
A44: j<=k1;
j <> len f+1 by A43,ZFMISC_1:56;
then jlen f+1;
set LA=Line(AA,k1);
set LAf=Line(AA,len f+1);
set a=AA*(len f+1,n1);
set RA=RLine(AA,k1,LA + (-(AA*(k1,n1))*a")*LAf);
A46: width AA=m() by A4,MATRIX_0:23;
then
A47: len (LA+(-(AA*(k1,n1))*a")*LAf)=m()by CARD_1:def 7;
A48: Indices A9=Indices AA by MATRIX_0:26;
then [len f+1,n1] in Indices AA by A25,A30,A26,ZFMISC_1:87;
then
A49: RA*(len f+1,n1)<>0.K() by A38,A45,A46,A47,MATRIX11:def 3;
A50: Indices A9=Indices RA by MATRIX_0:26;
A51: now
let i,j such that
A52: [i,j] in Indices RA and
A53: i>len f and
A54: j<=n;
now
per cases;
suppose
A55: i = k1;
A56: j in Seg m() by A25,A50,A52,ZFMISC_1:87;
then len f+1>len f & [len f+1,j] in Indices A9 by A25,A30
,NAT_1:13,ZFMISC_1:87;
then AA*(len f+1,j)=0.K() by A39,A48,A54;
then LAf.j=0.K() by A46,A56,MATRIX_0:def 7;
then
A57: ((-(AA*(k1,n1))*a")*LAf).j = (-(AA*(k1,n1))*a")*0.
K() by A46,A56,FVSUM_1:51
.= 0.K();
LA.j = AA*(k1,j) by A46,A56,MATRIX_0:def 7
.= 0.K() by A39,A48,A50,A52,A53,A54,A55;
then 0.K()+0.K() = (LA + (-(AA*(k1,n1))*a")*LAf).j by A46
,A56,A57,FVSUM_1:18
.= RA*(i,j) by A48,A50,A46,A47,A52,A55,MATRIX11:def 3;
hence RA*(i,j) = 0.K() by RLVECT_1:def 4;
end;
suppose
i<>k1;
hence RA*(i,j) = AA*(i,j) by A48,A50,A46,A47,A52,
MATRIX11:def 3
.= 0.K() by A39,A48,A50,A52,A53,A54;
end;
end;
hence RA*(i,j)=0.K();
end;
set RB=F(BB,k1,len f+1,-(AA*(k1,n1))*a");
take RA,RB;
A58: width RA=m() by A4,MATRIX_0:23;
A59: len AA=n() by MATRIX_0:def 2;
A60: now
A61: dom AA=Seg len AA by FINSEQ_1:def 3;
let j such that
A62: j in dom RA \ {len f+1} and
A63: j<=k1;
j in dom RA by A62,XBOOLE_0:def 5;
then
A64: [j,n1] in Indices AA by A26,A48,A50,A58,ZFMISC_1:87;
A65: dom RA=Seg len RA & len RA=n() by FINSEQ_1:def 3
,MATRIX_0:def 2;
now
per cases by A63,NAT_1:8;
suppose
A66: j<=k;
then jlen f & fi<=n by A14,A72,FINSEQ_1:1,NAT_1:13;
[len f+1,fi] in Indices AA & LAf.fi=AA*(len f+1,fi) by A25
,A32,A30,A48,A46,A73,MATRIX_0:def 7,ZFMISC_1:87;
then LAf.fi=0.K() by A39,A74;
then
A75: ((-(AA*(k1,n1))*a")*LAf).fi =(-(AA*(k1,n1))*a")*0.K()
by A32,A46,A73,FVSUM_1:51
.= 0.K();
LA.fi=AA*(k1,fi) by A32,A46,A73,MATRIX_0:def 7;
then
A76: (LA+(-(AA*(k1,n1))*a")*LAf).fi = AA*( k1,fi)+0.K() by A32,A46
,A73,A75,FVSUM_1:18
.= AA*(k1,fi) by RLVECT_1:def 4;
A77: [i,fi] in Indices AA by A25,A32,A33,A31,A48,A71,A73,
ZFMISC_1:87;
now
per cases;
suppose
i<>k1;
then RA*(i,fi)=AA*(i,fi)by A46,A47,A77,MATRIX11:def 3;
hence RA*(i,fi)<>0.K() by A40,A71;
end;
suppose
i=k1;
then RA*(i,fi)=AA*(i,fi) by A46,A47,A77,A76,
MATRIX11:def 3;
hence RA*(i,f/.i)<>0.K() by A40,A71;
end;
end;
hence RA*(i,fi) <> 0.K()& (j in dom f & i < j implies fi < f
/.j) by A15,A71;
thus j in dom f \ {i} implies RA*(j,fi) = 0.K()
proof
assume
A78: j in dom f \ {i};
then j in Seg len f by A31,XBOOLE_0:def 5;
then
A79: [j,fi] in Indices AA by A25,A32,A33,A48,A73,ZFMISC_1:87;
per cases;
suppose
j<>k1;
hence RA*(j,fi) = AA*(j,fi) by A46,A47,A79,MATRIX11:def 3
.= 0.K() by A40,A71,A78;
end;
suppose
A80: j = k1;
hence
RA*(j,fi) = (LA+(-(AA*(k1,n1))*a")*LAf).fi by A46,A47,A79
,MATRIX11:def 3
.= 0.K() by A40,A71,A76,A78,A80;
end;
end;
thus j in Seg width RA & j < f/.i implies RA*(i,j) = 0.K()
proof
assume that
A81: j in Seg width RA and
A82: j < f/.i;
A83: [len f+1,j] in Indices AA by A30,A59,A69,A46,A58,A81,
ZFMISC_1:87;
A84: [i,j] in Indices AA by A33,A31,A59,A69,A46,A58,A71,A81,
ZFMISC_1:87;
per cases;
suppose
i<>k1;
hence RA*(i,j) = AA*(i,j) by A46,A47,A84,MATRIX11:def 3
.= 0.K() by A40,A46,A58,A71,A81,A82;
end;
suppose
A85: i = k1;
fi <= n by A14,A72,FINSEQ_1:1;
then
A86: j <= n by A82,XXREAL_0:2;
len f+1 > len f by NAT_1:13;
then 0.K() = AA*(len f+1,j) by A39,A83,A86
.= LAf.j by A46,A58,A81,MATRIX_0:def 7;
then
A87: ((-(AA*(k1,n1))*a")*LAf).j = (-(AA*(k1,n1))*a")*0.
K() by A46,A58,A81,FVSUM_1:51
.= 0.K();
LA.j = AA*(i,j)by A46,A58,A81,A85,MATRIX_0:def 7
.= 0.K() by A40,A46,A58,A71,A81,A82;
then (LA + (-(AA*(k1,n1))*a")*LAf).j = 0.K()+0.K() by A46
,A58,A81,A87,FVSUM_1:18
.= 0.K() by RLVECT_1:def 4;
hence thesis by A46,A47,A84,A85,MATRIX11:def 3;
end;
end;
end;
A88: len f+1 in Seg len AA by A4,A30,MATRIX_0:23;
then
A89: r=the_rank_of RA by A37,A45,MATRIX13:92;
P[RA,RB] by A2,A36,A45,A88,A69;
hence thesis by A89,A49,A51,A70,A60;
end;
end;
hence thesis;
end;
A90: j0 in Seg m() by A21,A25,ZFMISC_1:87;
ex A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() st
P[A9,B9] & r = the_rank_of A9 & A9*(len f+1,n1) <> 0.K() & (for i,j st [i,j] in
Indices A9 & i>len f &j<=n holds A9*(i,j)=0.K())& for i,j st i in dom f holds
PP[f,i,j,A9]
proof
per cases;
suppose
A9*(len f+1,n1) <> 0.K();
hence thesis by A8,A10,A15;
end;
suppose
A91: A9*(len f+1,n1) = 0.K();
set RB=F(B9,len f+1,i0,1_K());
set LA=Line(A9,i0);
set LAf=Line(A9,len f+1);
set RA=RLine(A9,len f+1,LAf + 1_K()*LA);
take RA,RB;
i0 in dom A9 & dom A9=Seg len A9 by A21,FINSEQ_1:def 3
,ZFMISC_1:87;
hence P[RA,RB] & r = the_rank_of RA by A2,A8,A23,A24,A91,
MATRIX13:92;
A92: 1_K()*LA=LA & len (LAf+LA)=width A9 by CARD_1:def 7,FVSUM_1:57;
[len f+1,j0] in Indices A9 by A25,A30,A90,ZFMISC_1:87;
then
A93: RA*(len f+1,n1)=(LAf+LA).n1 by A23,A92,MATRIX11:def 3;
A94: width A9=m() by A4,MATRIX_0:23;
then
A95: LA.n1=A9*(i0,n1) by A23,A90,MATRIX_0:def 7;
LAf.n1=0.K() by A23,A90,A91,A94,MATRIX_0:def 7;
then RA*(len f+1,n1) = 0.K()+A9*(i0,n1) by A26,A94,A93,A95,
FVSUM_1:18
.= A9*(i0,n1) by RLVECT_1:def 4;
hence RA*(len f+1,n1)<>0.K() by A23,A24;
A96: Indices RA=Indices A9 by MATRIX_0:26;
now
let i,j such that
A97: [i,j] in Indices RA and
A98: i>len f and
A99: j<=n;
A100: j in Seg m() by A94,A96,A97,ZFMISC_1:87;
A101: i>=len f+1 by A98,NAT_1:13;
now
per cases by A101,XXREAL_0:1;
suppose
i>len f+1;
hence RA*(i,j) = A9*(i,j) by A92,A96,A97,MATRIX11:def 3
.= 0.K() by A10,A96,A97,A98,A99;
end;
suppose
A102: i=len f+1;
A103: [i0,j] in Indices A9 by A25,A27,A100,ZFMISC_1:87;
A104: LAf.j=A9*( len f+1,j) & LA.j=A9*(i0,j) by A94,A100,
MATRIX_0:def 7;
RA*(i,j)=(LAf+LA).j by A92,A96,A97,A102,MATRIX11:def 3;
hence RA*(i,j) = A9*(len f+1,j)+A9*(i0,j) by A94,A100,A104,
FVSUM_1:18
.= 0.K()+A9*(i0,j) by A10,A96,A97,A98,A99,A102
.= 0.K()+0.K() by A10,A22,A99,A103
.= 0.K() by RLVECT_1:def 4;
end;
end;
hence RA*(i,j)=0.K();
end;
hence
for i,j st [i,j] in Indices RA & i > len f&j<=n holds RA*(i
,j) = 0.K();
let i,j such that
A105: i in dom f;
i <= len f by A31,A105,FINSEQ_1:1;
then
A106: i0.K()&(j in dom f & i 0.K()
& for i,j st [ i,j] in Indices A0& i>len f& j<=n holds A0*(i,j)=0.K() ) & for i
,j st i in dom f holds PP[f,i,j,A0];
A115: QQ[0]
proof
assume 0<=n();
take A0,B0;
now
A116: dom A0=Seg len A0 by FINSEQ_1:def 3;
let j;
assume j in dom A0 \ {len f+1} & j<=0;
hence A0*(j,n1) = 0.K() by A116;
end;
hence thesis by A114;
end;
for k holds QQ[k] from NAT_1:sch 2(A115,A34);
then consider
Aa be Matrix of n(),m(),K(),Bb be Matrix of n(),m9(),K()
such that
A117: ( P[Aa,Bb])& r = the_rank_of Aa and
A118: Aa*(len f+1,n1) <> 0.K() and
A119: for i,j st [i,j] in Indices Aa & i>len f&j<=n holds Aa*(i,
j)=0.K() and
A120: for i,j st i in dom f holds PP[f,i,j,Aa] and
A121: for j st j in dom Aa\{len f+1}&j<=n() holds Aa*(j,n1)=0.K( );
take Aa,Bb;
thus P[Aa,Bb] & r = the_rank_of Aa by A117;
take f9=f^<*n1*>;
A122: len f9=len f+1 by FINSEQ_2:16;
A123: len Aa=n() & dom Aa=Seg len Aa by A4,FINSEQ_1:def 3,MATRIX_0:23;
A124: now
let i,j such that
A125: [i,j] in Indices Aa and
A126: i>len f9 and
A127: j<=n1;
per cases by A127,NAT_1:8;
suppose
A128: j<=n;
i>len f by A122,A126,NAT_1:13;
hence Aa*(i,j)=0.K() by A119,A125,A128;
end;
suppose
A129: j=n1;
i in dom Aa by A125,ZFMISC_1:87;
then i in dom Aa\{len f+1} & i<=n() by A122,A123,A126,FINSEQ_1:1
,ZFMISC_1:56;
hence Aa*(i,j)=0.K() by A121,A129;
end;
end;
A130: width Aa=m() by A4,MATRIX_0:23;
A131: len f9<=n() by A28,A29,A122,XXREAL_0:2;
A132: now
let i,j such that
A133: i in dom f9;
A134: dom f9 = Seg (len f+1) by A122,FINSEQ_1:def 3
.= dom f\/{len f+1} by A31,FINSEQ_1:9;
A135: now
let k such that
A136: k in dom f;
A137: k in dom f9 by A134,A136,XBOOLE_0:def 3;
thus f/.k = f.k by A136,PARTFUN1:def 6
.= f9.k by A136,FINSEQ_1:def 7
.= f9/.k by A137,PARTFUN1:def 6;
end;
now
per cases by A133,A134,XBOOLE_0:def 3;
suppose
A138: i in dom f;
then f/.i=f9/.i by A135;
hence Aa*(i,f9/.i)<>0.K() by A120,A138;
end;
suppose
A139: i in {len f+1};
A140: f9/.i=f9.i by A133,PARTFUN1:def 6;
i=len f+1 by A139,TARSKI:def 1;
hence Aa*(i,f9/.i)<>0.K() by A118,A140,FINSEQ_1:42;
end;
end;
hence Aa*(i,f9/.i) <> 0.K();
thus j in dom f9 & i < j implies f9/.i < f9/.j
proof
assume that
A141: j in dom f9 and
A142: i < j;
per cases by A133,A134,A141,XBOOLE_0:def 3;
suppose
A143: j in {len f+1} & i in {len f+1};
then i=len f+1 by TARSKI:def 1;
hence thesis by A142,A143,TARSKI:def 1;
end;
suppose
A144: j in {len f+1} & i in dom f;
then len f+1=j by TARSKI:def 1;
then
A145: f9.j=n1 by FINSEQ_1:42;
f/.i=f.i & f.i in rng f by A144,FUNCT_1:def 3,PARTFUN1:def 6;
then
A146: f/.i <= n by A14,FINSEQ_1:1;
f9.j=f9/.j by A141,PARTFUN1:def 6;
then f/.i < f9/.j by A146,A145,NAT_1:13;
hence thesis by A135,A144;
end;
suppose
j in dom f & i in {len f+1};
then j<=len f & i=len f+1 by A31,FINSEQ_1:1,TARSKI:def 1;
hence thesis by A142,NAT_1:13;
end;
suppose
A147: j in dom f & i in dom f;
then f/.i=f9/.i & f/.j=f9/.j by A135;
hence thesis by A15,A142,A147;
end;
end;
dom f9=Seg len f9 by FINSEQ_1:def 3;
then
A148: dom f9 c= dom Aa by A123,A131,FINSEQ_1:5;
thus j in dom f9 \ {i} implies Aa*(j,f9/.i) = 0.K()
proof
assume
A149: j in dom f9 \ {i};
per cases;
suppose
A150: i=len f+1;
len f+1 in {len f+1} by TARSKI:def 1;
then len f+1 in dom f9 by A134,XBOOLE_0:def 3;
then
A151: f9.(len f+1)=f9/.i by A150,PARTFUN1:def 6;
A152: j in dom f9 by A149,ZFMISC_1:56;
j<>i by A149,ZFMISC_1:56;
then
A153: j in dom Aa\{i} by A148,A152,ZFMISC_1:56;
j<=n() by A123,A148,A152,FINSEQ_1:1;
then Aa*(j,n1)=0.K() by A121,A150,A153;
hence thesis by A151,FINSEQ_1:42;
end;
suppose
A154: i<>len f+1;
A155: j in dom f9 by A149,XBOOLE_0:def 5;
A156: i in dom f or i in {len f+1} by A133,A134,XBOOLE_0:def 3;
then
A157: f.i in rng f by A154,FUNCT_1:def 3,TARSKI:def 1;
A158: f/.i=f9/.i & f.i=f/.i by A135,A154,A156,PARTFUN1:def 6
,TARSKI:def 1;
then
A159: 1<=f9/.i by A14,A157,FINSEQ_1:1;
A160: f9/.i<=n by A14,A158,A157,FINSEQ_1:1;
n<=m() by A7,NAT_1:13;
then f9/.i <=m() by A160,XXREAL_0:2;
then f9/.i in Seg width Aa by A130,A159;
then
A161: [j,f9/.i] in Indices Aa by A148,A155,ZFMISC_1:87;
per cases;
suppose
j=len f+1;
then j > len f by NAT_1:13;
hence thesis by A119,A160,A161;
end;
suppose
A162: j<>len f+1;
j in dom f9 by A149,XBOOLE_0:def 5;
then
A163: j in dom f or j in {len f+1} by A134,XBOOLE_0:def 3;
j<>i by A149,ZFMISC_1:56;
then j in dom f\{i} by A162,A163,TARSKI:def 1,ZFMISC_1:56;
then Aa*(j,f/.i)=0.K() by A120,A154,A156,TARSKI:def 1;
hence thesis by A135,A154,A156,TARSKI:def 1;
end;
end;
end;
thus j in Seg width Aa & j < f9/.i implies Aa*(i,j) = 0.K()
proof
assume that
A164: j in Seg width Aa and
A165: j < f9/.i;
per cases;
suppose
A166: i in dom f;
then f9/.i=f/.i by A135;
hence thesis by A120,A164,A165,A166;
end;
suppose
A167: not i in dom f;
i in dom f or i in {len f+1} by A133,A134,XBOOLE_0:def 3;
then
A168: i=len f+1 by A167,TARSKI:def 1;
then
A169: f9.i = n1 by FINSEQ_1:42;
f9.i=f9/.i by A133,PARTFUN1:def 6;
then
A170: j<=n by A165,A169,NAT_1:13;
A171: i>len f by A168,NAT_1:13;
[i,j] in Indices Aa by A133,A148,A164,ZFMISC_1:87;
hence thesis by A119,A170,A171;
end;
end;
end;
A172: rng <*n1*>= {n1} & rng f \/{n1} c= Seg n\/{n1} by A14,FINSEQ_1:38
,XBOOLE_1:9;
A173: Seg n\/{n1}=Seg n1 by FINSEQ_1:9;
rng f misses {n1} & <*n1*> is one-to-one by A14,FINSEQ_3:14
,XBOOLE_1:63;
hence thesis by A11,A12,A28,A29,A122,A124,A172,A173,A132,FINSEQ_1:31
,FINSEQ_3:91,XREAL_1:6,XXREAL_0:2;
end;
end;
A174: Q[0]
proof
assume 0<=m();
take A9=A(),B9=B();
thus P[A9,B9] & r = the_rank_of A9 by A1;
take f=<*>NAT;
now
let i,j such that
A175: [i,j] in Indices A9 and
i > len f and
A176: j<=0;
j in Seg width A9 by A175,ZFMISC_1:87;
hence A9*(i,j)=0.K() by A176;
end;
hence thesis;
end;
for n holds Q[n] from NAT_1:sch 2(A174,A5);
then consider
A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() such
that
A177: P[A9,B9] and
A178: r = the_rank_of A9 and
A179: ex f be FinSequence of NAT st (for i,j st [i,j] in Indices A9&
i>len f&j<=m() holds A9*(i,j)=0.K())& f is one-to-one & len f <= m() & len f<=n
() & rng f c= Seg m() & for i,j st i in dom f holds PP[f,i,j,A9];
consider f be FinSequence of NAT such that
A180: for i,j st [i,j] in Indices A9&i>len f&j<=m() holds A9*(i,j)= 0.K() and
A181: f is one-to-one and
len f <= m() and
A182: len f<=n() and
A183: rng f c= Seg m() and
A184: for i,j st i in dom f holds PP[f,i,j,A9] by A179;
A185: len A9=n() by A4,MATRIX_0:23;
take A9,B9;
thus P[A9,B9] & r=the_rank_of A9 by A177,A178;
A186: dom A9=Seg len A9 by FINSEQ_1:def 3;
set L=len f;
A187: Seg L c= Seg n() by A182,FINSEQ_1:5;
idseq L is FinSequence of NAT & len (idseq L)=L by CARD_1:def 7
,FINSEQ_2:48;
then reconsider idL=idseq L,F9=f as Element of L-tuples_on NAT by
FINSEQ_2:92;
set S=Segm(A9,idL,F9);
A188: dom f=Seg L by FINSEQ_1:def 3;
set D=diagonal_of_Matrix S;
A189: Indices S=[:Seg L,Seg L:] by MATRIX_0:24;
for k be Nat st k in dom D holds D.k <> 0.K()
proof
A190: len D=L by MATRIX_3:def 10;
let k be Nat;
assume k in dom D;
then
A191: k in Seg L by A190,FINSEQ_1:def 3;
then [k,k] in Indices S by A189,ZFMISC_1:87;
then S*(k,k) = A9*(idL.k,f.k) by MATRIX13:def 1
.= A9*(k,f.k) by A191,FINSEQ_2:49
.= A9*(k,f/.k) by A188,A191,PARTFUN1:def 6;
then S*(k,k)<>0.K() by A184,A188,A191;
hence thesis by A191,MATRIX_3:def 10;
end;
then
A192: Product D<>0.K() by FVSUM_1:82;
now
let i,j such that
A193: [i,j] in Indices S;
A194: i in Seg L by A189,A193,ZFMISC_1:87;
assume i>j;
then
A195: i in dom f \{j} by A188,A194,ZFMISC_1:56;
reconsider i9=i,j9=j as Element of NAT by ORDINAL1:def 12;
A196: j in Seg L by A189,A193,ZFMISC_1:87;
thus S*(i,j) = A9*(idL.i9,f.j9) by A193,MATRIX13:def 1
.= A9*(i,f.j) by A194,FINSEQ_2:49
.= A9*(i,f/.j) by A188,A196,PARTFUN1:def 6
.= 0.K() by A184,A188,A196,A195;
end;
then S is upper_triangular Matrix of L,K() by MATRIX_1:def 8;
then
A197: Det S<>0.K() by A192,MATRIX13:7;
A198: len Segm(A9,Seg L,Seg width A9)=card Seg L by MATRIX_0:def 2;
A199: width A9=m() by A4,MATRIX_0:23;
[:rng idL,rng F9:] c= Indices A9 by A183,A187,A185,A186,A199,
ZFMISC_1:96;
then
A200: the_rank_of A9 >= L by A197,MATRIX13:76;
A201: now
set w0=width A9 |-> 0.K();
let i such that
A202: i in dom A9\(Seg L);
set LA=Line(A9,i);
A203: now
not i in Seg L by A202,XBOOLE_0:def 5;
then
A204: i > L or i<1;
let j such that
A205: 1<=j and
A206: j<=width A9;
A207: j in Seg width A9 by A205,A206;
A208: i in dom A9 by A202,XBOOLE_0:def 5;
then
A209: [i,j] in Indices A9 by A207,ZFMISC_1:87;
thus LA.j = A9*(i,j) by A207,MATRIX_0:def 7
.= 0.K() by A180,A186,A199,A206,A208,A209,A204,FINSEQ_1:1
.= w0.j by A207,FINSEQ_2:57;
end;
len LA=width A9 & len w0=width A9 by CARD_1:def 7;
hence Line(A9,i) = width A9 |-> 0.K() by A203;
end;
then the_rank_of A9=the_rank_of Segm(A9,Seg L,Seg width A9) by A187,A185
,A186,Th11;
then the_rank_of A9 <= card Seg L by A198,MATRIX13:74;
then
A210: the_rank_of A9 <= L by FINSEQ_1:57;
then
A211: r=L by A178,A200,XXREAL_0:1;
thus for i st i in dom A9 & i > r holds Line(A9,i)=m() |->0.K()
proof
let i such that
A212: i in dom A9 and
A213: i > r;
not i in Seg L by A211,A213,FINSEQ_1:1;
then i in dom A9\(Seg L) by A212,XBOOLE_0:def 5;
hence thesis by A199,A201;
end;
take f;
thus thesis by A4,A181,A183,A184,A200,A210,MATRIX_0:23,XXREAL_0:1;
end;
end;
then consider
A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() such that
A214: P[A9,B9] and
A215: r = the_rank_of A9 and
A216: for i st i in dom A9 & i > r holds Line(A9,i)=m() |->0.K() and
A217: ex f be FinSequence of NAT st len f = the_rank_of A9 & f is
one-to-one & rng f c= Seg width A9 & for i,j st i in dom f holds PP[f,i,j,A9]
;
consider f be FinSequence of NAT such that
A218: len f = the_rank_of A9 and
A219: f is one-to-one and
A220: rng f c= Seg width A9 and
A221: for i,j st i in dom f holds PP[f,i,j,A9] by A217;
not 0 in rng f by A220;
then reconsider rngf=rng f as without_zero finite Subset of NAT by A220,
MEASURE6:def 2,XBOOLE_1:1;
A222: n()=0 or n()>0;
set S=Segm(A9,Seg card rngf,rngf);
A223: dom f=Seg r by A215,A218,FINSEQ_1:def 3;
take A9,B9,rngf;
len A9 = n() by MATRIX_0:def 2;
then width A9 = 0 or width A9 = m() by A222,MATRIX_0:23,def 3;
then Seg width A9 c= Seg m();
hence rngf c= Seg m() & r = the_rank_of A9 by A215,A220;
dom f,rngf are_equipotent by A219,WELLORD2:def 4;
hence
A224: card rngf = card dom f by CARD_1:5
.= card Seg r by A215,A218,FINSEQ_1:def 3
.= r by FINSEQ_1:57;
now
let i,j;
assume that
A225: i in dom f & j in dom f and
A226: i < j;
f.i=f/.i & f.j=f/.j by A225,PARTFUN1:def 6;
hence f.i < f.j by A221,A225,A226;
end;
then
A227: Sgm rngf = f by A219,A220,Th6;
thus P[A9,B9] by A214;
A228: card Seg card rngf=card rngf by FINSEQ_1:57;
then
A229: Indices S = [:Seg r,Seg r:] by A224,MATRIX_0:24;
now
let i,j be Nat such that
A230: i in Seg r and
A231: j in Seg r and
A232: i <> j;
A233: i in dom f\{j} by A223,A230,A232,ZFMISC_1:56;
A234: idseq r.i=i by A230,FINSEQ_2:49;
[i,j] in Indices S by A229,A230,A231,ZFMISC_1:87;
then S*(i,j) = A9*(Sgm (Seg r).i,f.j) by A224,A227,MATRIX13:def 1
.= A9*(i,f.j) by A234,FINSEQ_3:48
.= A9*(i,f/.j) by A223,A231,PARTFUN1:def 6;
hence S*(i,j) = 0.K() by A221,A223,A231,A233;
end;
hence S is diagonal by A224,A228,MATRIX_7:def 2;
thus for i st i in Seg card rngf holds A9*(i,Sgm rngf/.i) <> 0.K() by A221
,A224,A227,A223;
thus for i st i in dom A9 & i > card rngf holds Line(A9,i) = m() |->0.K() by
A216,A224;
let i,j such that
A235: i in Seg card rngf and
A236: j in Seg width A9 and
A237: j < Sgm rngf.i;
j Field, n,m,m9() -> Nat, A() -> Matrix of n(),m(),K(), B() ->
Matrix of n(),m9(),K(), F(Matrix of n(),m9(),K(),Nat,Nat,Element of K()) ->
Matrix of n(),m9(),K(), P[set,set]}: ex A9 be Matrix of n(),m(),K(), B9 be
Matrix of n(),m9(),K(), N be without_zero finite Subset of NAT st N c= Seg m()
& the_rank_of A() = the_rank_of A9 & the_rank_of A() = card N & P[A9,B9] & Segm
(A9,Seg card N,N) = 1.(K(),card N) & ( for i st i in dom A9 & i > card N holds
Line(A9,i) = m() |->0.K() ) & for i,j st i in Seg card N & j in Seg width A9 &
j < Sgm N.i holds A9*(i,j) = 0.K()
provided
A1: P[A(),B()] and
A2: for A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() st P[
A9,B9] for a be Element of K() for i,j st j in dom A9 & (i = j implies a <> -
1_K() ) holds P[RLine(A9,i,Line(A9,i) + a*Line(A9,j)),F(B9,i,j,a)]
proof
set r=the_rank_of A();
A3: for A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K() st P[A9,B9
] for i,j st i <> j & j in dom A9 for a be Element of K() holds P[RLine(A9,i,
Line(A9,i) + a*Line(A9,j)),F(B9,i,j,a)] by A2;
consider A9 be Matrix of n(),m(),K(), B9 be Matrix of n(),m9(),K(), N be
without_zero finite Subset of NAT such that
A4: N c= Seg m() and
A5: r = the_rank_of A9 & r = card N and
A6: ( P[A9,B9] & Segm(A9,Seg card N,N) is diagonal & for i st i in Seg
card N holds A9*(i,Sgm N/.i) <> 0.K() )&( ( for i st i in dom A9 & i > card N
holds Line(A9,i) = m() |->0.K())& for i,j st i in Seg card N &
j in Seg width A9
& j < Sgm N.i holds A9*(i,j) = 0.K() ) from GAUSS1(A1,A3);
set ONE=1.(K(),card N);
A7: Indices ONE = [:Seg r,Seg r:] by A5,MATRIX_0:24;
defpred Q[Nat] means $1 <= card N implies ex A9 be Matrix of n(),m(),K(), B9
be Matrix of n(),m9(),K() st r = the_rank_of A9 & ( for i st i in Seg card N &
i<=$1 holds A9*(i,Sgm N/.i) = 1_K() )& P[A9,B9] & Segm(A9,Seg card N,N) is
diagonal & ( for i st i in Seg card N holds A9*(i,Sgm N/.i) <> 0.K() )& ( for i
st i in dom A9 & i > card N holds Line(A9,i) = m() |->0.K())& ( for i,j st i in
Seg card N & j in Seg width A9 & j < Sgm N.i holds A9*(i,j) = 0.K());
A8: for n st Q[n] holds Q[n+1]
proof
set f=Sgm N;
let n such that
A9: Q[n];
set n1=n+1;
assume
A10: n1<=card N;
then consider
A1 be Matrix of n(),m(),K(),A25 be Matrix of n(),m9(),K() such
that
A11: r = the_rank_of A1 and
A12: for i st i in Seg card N & i<=n holds A1*(i,Sgm N/.i) = 1_K() and
A13: P[A1,A25] and
A14: Segm(A1,Seg card N,N) is diagonal and
A15: for i st i in Seg card N holds A1*(i,Sgm N/.i) <> 0.K() and
A16: for i st i in dom A1 & i > card N holds Line(A1,i) = m() |->0.K() and
A17: for i,j st i in Seg card N & j in Seg width A1 & j < Sgm N.i
holds A1*(i,j) = 0.K() by A9,NAT_1:13;
set L=Line(A1,n1);
set LL=L+((A1*(n1,f/.n1))"-1_K())*L;
set R=RLine(A1,n1,LL);
take R,FB=F(A25,n1,n1,(A1*(n1,f/.n1))"-1_K());
A18: len A1=n() by MATRIX_0:def 2;
set SA=Segm(A1,Seg card N,N);
set S=Segm(R,Seg card N,N);
A19: len LL=width A1 by CARD_1:def 7;
r<=len A9 & len A9=n() by A5,MATRIX13:74,MATRIX_0:def 2;
then
A20: Seg r c= Seg n() by FINSEQ_1:5;
1<=1+n by NAT_1:11;
then
A21: n1 in Seg card N by A10;
then
A22: n()<>0 by A5,A20;
then
A23: width R=m() by MATRIX_0:23;
A24: (A1*(n1,Sgm N/.n1))"-1_K()<>-1.K()
proof
assume
A25: (A1*(n1,Sgm N/.n1))"-1_K()=-1.K();
A26: 0.K() = 1_K()+ -1_K() by VECTSP_1:19
.= (1_K()+(-1_K()))+A1*(n1,Sgm N/.n1)" by A25,RLVECT_1:def 3
.= 0.K()+A1*(n1,Sgm N/.n1)" by VECTSP_1:19
.= A1*(n1,Sgm N/.n1)" by RLVECT_1:def 4;
A1*(n1,Sgm N/.n1)<>0.K() by A15,A21;
hence thesis by A26,VECTSP_1:25;
end;
hence the_rank_of R =r by A5,A11,A21,A20,A18,MATRIX13:92;
A27: width A1=m() by A22,MATRIX_0:23;
A28: LL = 1_K()*L+((A1*(n1,f/.n1))"-1_K())*L by FVSUM_1:57
.= (1_K() +(-1_K()+(A1*(n1,f/.n1))"))*L by FVSUM_1:55
.= ((1_K() +-1_K())+(A1*(n1,f/.n1))")*L by RLVECT_1:def 3
.= (0.K()+A1*(n1,f/.n1)")*L by VECTSP_1:19
.= (A1*(n1,f/.n1))"*L by RLVECT_1:def 4;
A29: dom A1=Seg len A1 by FINSEQ_1:def 3;
A30: rng Sgm N=N by A4,FINSEQ_1:def 13;
A31: dom Sgm N=Seg card N by A4,FINSEQ_3:40;
thus
A32: for i st i in Seg card N & i<=n1 holds R*(i,Sgm N/.i) = 1_K()
proof
let i;
assume that
A33: i in Seg card N and
A34: i<=n1;
A35: f.i=f/.i & f.i in rng f by A31,A33,FUNCT_1:def 3,PARTFUN1:def 6;
then
A36: [i,Sgm N/.i] in Indices A1 by A4,A5,A30,A20,A29,A18,A27,A33,ZFMISC_1:87;
per cases by A34,NAT_1:8;
suppose
A37: i<=n;
then i0.K() by A15,A33;
R*(i,f/.i) = ((A1*(i,f/.i))"*L).(f/.i) & L.(f/.i)=A1*(i,f/.i) by A4,A30
,A19,A28,A27,A35,A36,A38,MATRIX11:def 3,MATRIX_0:def 7;
hence R*(i,f/.i) = (A1*(i,f/.i))"*(A1*(i,f/.i)) by A4,A30,A27,A35,
FVSUM_1:51
.= 1_K() by A39,VECTSP_1:def 10;
end;
end;
thus P[R,FB] by A2,A5,A13,A21,A20,A29,A18,A24;
thus S is diagonal
proof
A40: Indices A1=Indices R by MATRIX_0:26;
let i,j such that
A41: [i,j] in Indices S and
A42: S*(i,j) <> 0.K();
reconsider I=i,J=j as Element of NAT by ORDINAL1:def 12;
Indices S=[:Seg card Seg card N,Seg width S:] by MATRIX_0:25;
then i in Seg card Seg card N by A41,ZFMISC_1:87;
then i in Seg card card N by FINSEQ_1:55;
then
A43: Sgm (Seg card N)=idseq card N & (idseq card N).I=I by FINSEQ_2:49
,FINSEQ_3:48;
then
A44: S*(i,j)=R*(I,Sgm N.J) by A41,MATRIX13:def 1;
rng Sgm (Seg card N)=(Seg card N) & [:Seg card N,N:] c= Indices A1
by A4,A5,A20,A29,A18,A27,FINSEQ_1:def 13,ZFMISC_1:96;
then
A45: [I,Sgm N.J] in Indices A1 by A30,A41,A40,A43,MATRIX13:17;
then
A46: Sgm N.J in Seg width A1 by ZFMISC_1:87;
then reconsider SgmNJ=Sgm N.j as Element of NAT;
A47: Indices S=Indices SA by MATRIX_0:26;
per cases;
suppose
A48: I=n1;
thus i=j
proof
assume
A49: i<>j;
R*(I,Sgm N.J)=((A1*(n1,f/.n1))"*L).SgmNJ & L.SgmNJ=A1*(I,SgmNJ)
by A19,A28,A45,A46,A48,MATRIX11:def 3,MATRIX_0:def 7;
then R*(I,Sgm N.J) = (A1*(n1,f/.n1))"*(A1*(I,SgmNJ)) by A46,
FVSUM_1:51
.= (A1*(n1,f/.n1))"*(SA*(i,j)) by A41,A47,A43,MATRIX13:def 1
.= (A1*(n1,f/.n1))"*(0.K()) by A14,A41,A47,A49
.= 0.K();
hence thesis by A41,A42,A43,MATRIX13:def 1;
end;
end;
suppose
I<>n1;
then R*(I,Sgm N.J) = A1*(Sgm (Seg card N).I,Sgm N.J) by A19,A43,A45,
MATRIX11:def 3
.= SA*(i,j) by A41,A47,MATRIX13:def 1;
hence thesis by A14,A41,A42,A47,A44;
end;
end;
thus for i st i in Seg card N holds R*(i,Sgm N/.i) <> 0.K()
proof
let i such that
A50: i in Seg card N;
f.i=f/.i & f.i in rng f by A31,A50,FUNCT_1:def 3,PARTFUN1:def 6;
then
A51: [i,Sgm N/.i] in Indices A1 by A4,A5,A30,A20,A29,A18,A27,A50,ZFMISC_1:87;
per cases;
suppose
i=n1;
hence thesis by A32,A50;
end;
suppose
i<>n1;
then R*(i,Sgm N/.i)=A1*(i,Sgm N/.i) by A19,A51,MATRIX11:def 3;
hence thesis by A15,A50;
end;
end;
thus for i st i in dom R & i > card N holds Line(R,i) = m() |->0.K()
proof
A52: dom R=Seg len R & len R=n() by FINSEQ_1:def 3,MATRIX_0:def 2;
let i such that
A53: i in dom R & i > card N;
thus Line(R,i) = Line(A1,i) by A10,A53,A52,MATRIX11:28
.= m() |->0.K() by A16,A29,A18,A53,A52;
end;
let i,j such that
A54: i in Seg card N and
A55: j in Seg width R and
A56: j < Sgm N.i;
A57: [i,j] in Indices A1 by A5,A20,A29,A18,A27,A23,A54,A55,ZFMISC_1:87;
per cases;
suppose
i=n1;
then R*(i,j)=((A1*(n1,f/.n1))"*L).j & L.j=A1*(i,j) by A19,A28,A27,A23,A55
,A57,MATRIX11:def 3,MATRIX_0:def 7;
hence R*(i,j) = (A1*(n1,f/.n1))"*(A1*(i,j)) by A27,A23,A55,FVSUM_1:51
.= (A1*(n1,f/.n1))"*(0.K()) by A17,A27,A23,A54,A55,A56
.= 0.K();
end;
suppose
i<>n1;
hence R*(i,j) = A1*(i,j)by A19,A57,MATRIX11:def 3
.= 0.K() by A17,A27,A23,A54,A55,A56;
end;
end;
for i st i in Seg card N & i<=0 holds A9*(i,Sgm N/.i) = 1_K();
then
A58: Q[0] by A5,A6;
for n holds Q[n] from NAT_1:sch 2(A58,A8);
then consider
A be Matrix of n(),m(),K(),B be Matrix of n(),m9(),K()such that
A59: r = the_rank_of A and
A60: for i st i in Seg card N & i<=card N holds A*(i,Sgm N/.i) = 1_K() and
A61: P[A,B] and
A62: Segm(A,Seg card N,N) is diagonal and
for i st i in Seg card N holds A*(i,Sgm N/.i) <> 0.K() and
A63: ( for i st i in dom A & i > card N holds Line(A,i) = m() |->0.K())&
for i,j st i in Seg card N & j in Seg width A & j < Sgm N.i holds A*(i,j) = 0.K
();
take A,B,N;
thus N c= Seg m() & r = the_rank_of A & r = card N & P[A,B] by A4,A5,A59,A61;
set S=Segm(A,Seg card N,N);
A64: card Seg card N=card N by FINSEQ_1:57;
then
A65: Indices ONE = Indices S by MATRIX_0:26;
now
A66: dom Sgm N=Seg r by A4,A5,FINSEQ_3:40;
let i,j such that
A67: [i,j] in Indices ONE;
A68: j in Seg r by A7,A67,ZFMISC_1:87;
reconsider i9=i,j9=j as Element of NAT by ORDINAL1:def 12;
A69: i in Seg r by A7,A67,ZFMISC_1:87;
then
A70: idseq r.i9=i9 by FINSEQ_2:49;
A71: i<=r by A69,FINSEQ_1:1;
A72: S*(i,j) = A*(Sgm (Seg r).i9,Sgm N.j9) by A5,A65,A67,MATRIX13:def 1
.= A*(i9,Sgm N.j9) by A70,FINSEQ_3:48
.= A*(i9,Sgm N/.j9) by A68,A66,PARTFUN1:def 6;
now
per cases;
suppose
A73: i9=j9;
hence S*(i,j) = 1_K() by A5,A60,A69,A71,A72
.= ONE*(i,j) by A67,A73,MATRIX_1:def 3;
end;
suppose
A74: i9<>j9;
hence S*(i,j) = 0.K() by A62,A65,A67
.= ONE*(i,j) by A67,A74,MATRIX_1:def 3;
end;
end;
hence S*(i,j) = ONE*(i,j);
end;
hence thesis by A63,A64,MATRIX_0:27;
end;
begin :: The main theorem
:: The Kronecker-Capelli theorem
theorem Th57:
for A,B be Matrix of K st len A = len B & (width A = 0 implies
width B = 0) holds the_rank_of A = the_rank_of (A^^B) iff Solutions_of(A,B) is
non empty
proof
let A,B be Matrix of K such that
A1: len A = len B and
A2: width A = 0 implies width B = 0;
set wB=width B;
set L=len A;
reconsider B9=B as Matrix of L,wB,K by A1,MATRIX_0:51;
set wA=width A;
reconsider A9=A as Matrix of L,wA,K by MATRIX_0:51;
deffunc F(Matrix of L,wB,K,Nat,Nat,Element of K) = RLine($1,$2,Line($1,$2) +
$4*Line($1,$3));
defpred P[set,set] means for A1 be Matrix of L,wA,K, B1 be Matrix of L,wB,K
st A1=$1 & B1=$2 holds the_rank_of (A9^^B9)=the_rank_of (A1^^B1) & Solutions_of
(A9,B9)=Solutions_of(A1,B1);
A3: for A1 be Matrix of L,wA,K, B1 be Matrix of L,wB,K st P[A1,B1] for a be
Element of K for i,j st j in dom A1 & (i = j implies a <> -1_K ) holds P[RLine(
A1,i,Line(A1,i) + a*Line(A1,j)),F(B1,i,j,a)]
proof
let A1 be Matrix of L,wA,K, B1 be Matrix of L,wB,K such that
A4: P[A1,B1];
let a be Element of K;
let i,j such that
A5: j in dom A1 and
A6: i = j implies a <> -1_K;
set LAj=Line(A1,j);
set LAi=Line(A1,i);
set RA=RLine(A1,i,LAi + a*LAj);
A7: dom A1 = Seg len A1 by FINSEQ_1:def 3
.= Seg L by MATRIX_0:def 2;
then
A8: Solutions_of(A1,B1)=Solutions_of(RA,F(B1,i,j,a)) by A5,A6,Th40;
set RB=F(B1,i,j,a);
set LBj=Line(B1,j);
set LBi=Line(B1,i);
A9: len A1=L & len B1=L by MATRIX_0:def 2;
per cases;
suppose
not i in Seg L;
then RA=A1 & RB=B1 by A9,MATRIX13:40;
hence thesis by A4;
end;
suppose
A10: i in Seg L;
A11: len (A1^^B1)=L by MATRIX_0:def 2;
A12: len (a*LAj)=width A1 & len (a*LBj)=width B1 by CARD_1:def 7;
A13: len LAi=width A1 & len LBi=width B1 by CARD_1:def 7;
len (LAi+a*LAj)=width A1 & len (LBi+a*LBj)=width B1 by CARD_1:def 7;
then RA^^RB = RLine(A1^^B1,i,(LAi+a*LAj)^(LBi+a*LBj))by Th18
.= RLine(A1^^B1,i,(LAi^LBi)+((a*LAj)^(a*LBj))) by A13,A12,Th3
.= RLine(A1^^B1,i,(LAi^LBi)+a*(LAj^LBj)) by Th4
.= RLine(A1^^B1,i,Line(A1^^B1,i)+a*(LAj^LBj))by A10,Th15
.= RLine(A1^^B1,i,Line(A1^^B1,i)+a*Line(A1^^B1,j)) by A5,A7,Th15;
then the_rank_of (RA^^RB)=the_rank_of (A1^^B1) by A5,A6,A7,A11,
MATRIX13:92;
hence thesis by A4,A8;
end;
end;
A14: P[A9,B9];
consider A1 be Matrix of L,wA,K, B1 be Matrix of L,wB,K,N such that
A15: N c= Seg wA and
A16: the_rank_of A9 = the_rank_of A1 & the_rank_of A9 = card N and
A17: P[A1,B1] & Segm(A1,Seg card N,N) = 1.(K,card N) and
A18: for i st i in dom A1 & i > card N holds Line(A1,i) = wA|->0.K and
for i,j st i in Seg card N & j in Seg width A1 & j < Sgm N.i holds A1*(i
,j) = 0.K from GAUSS2(A14,A3);
per cases;
suppose
A19: L=0;
then len (A9^^B9)=0 & the_rank_of A =0 by MATRIX13:74,MATRIX_0:def 2;
hence thesis by A19,Th51,MATRIX13:74;
end;
suppose
A20: L>0;
per cases;
suppose
A21: N<>{};
set SN=Seg card N;
set SS=Seg L\SN;
A22: card SN=card N by FINSEQ_1:57;
reconsider P2=Sgm SN " SN,Q2 = Sgm Seg wA " N as without_zero finite
Subset of NAT by MATRIX13:53;
dom Sgm SN=Seg card SN & rng Sgm SN=SN by FINSEQ_1:def 13,FINSEQ_3:40;
then
A23: P2 = SN by A22,RELAT_1:134;
rng Sgm Seg wA=Seg wA by FINSEQ_1:def 13;
then
A24: Sgm Seg wA.:Q2 = N by A15,FUNCT_1:77;
Q2 c= dom Sgm Seg wA & Sgm Seg wA is one-to-one by FINSEQ_3:92
,RELAT_1:132;
then N,Q2 are_equipotent by A24,CARD_1:33;
then
A25: dom Sgm Seg wA=Seg card Seg wA & card N=card Q2 by CARD_1:5,FINSEQ_3:40;
A26: Seg len A1=dom A1 & len A1=L by A20,FINSEQ_1:def 3,MATRIX_0:23;
A27: width A1=wA by A20,MATRIX_0:23;
A28: now
let i such that
A29: i in SS;
not i in SN by A29,XBOOLE_0:def 5;
then
A30: i<1 or i> card N;
i in Seg L by A29,XBOOLE_0:def 5;
hence Line(A1,i)=width A1|->0.K by A18,A26,A27,A30,FINSEQ_1:1;
end;
card N<=L by A16,MATRIX13:74;
then
A31: Seg card N c= Seg L by FINSEQ_1:5;
A32: len B1=L by A20,MATRIX_0:23;
A33: Seg len B1=dom B1 & width B1=wB by A20,FINSEQ_1:def 3,MATRIX_0:23;
thus the_rank_of A=the_rank_of(A^^B) implies Solutions_of(A,B) is non
empty
proof
assume the_rank_of A = the_rank_of (A^^B);
then the_rank_of A1 = the_rank_of (A1^^B1) by A16,A17;
then for i st i in (dom A1)\Seg card N holds Line(A1,i) = width A1|->
0.K & Line(B1,i) = width B1 |-> 0.K by A26,A32,A28,Th24,XBOOLE_1:36;
then
A34: Solutions_of(A1,B1)=Solutions_of(Segm(A1,SN,Seg wA), Segm(B1,SN,
Seg wB)) by A21,A31,A26,A32,A27,A33,Th45;
Segm(Segm(A1,SN,Seg wA),P2,Q2)=1.(K,card N) by A15,A17,MATRIX13:56;
then ex X be Matrix of card Seg wA,card Seg wB,K st Segm(X,Seg card
Seg wA \ Q2,Seg card Seg wB) = 0.(K,card Seg wA-'card SN,card Seg wB) & Segm(X,
Q2,Seg card Seg wB) = Segm(B1,SN,Seg wB) & X in Solutions_of(Segm(A1,SN,Seg wA)
,Segm(B1,SN,Seg wB)) by A21,A22,A23,A25,Th50,RELAT_1:132;
hence thesis by A17,A34;
end;
A35: Seg L\Seg card N c= Seg L by XBOOLE_1:36;
thus Solutions_of(A,B) is non empty implies the_rank_of A=the_rank_of (A
^^B)
proof
assume Solutions_of(A,B) is non empty;
then Solutions_of(A1,B1) is non empty by A17;
then consider x being object such that
A36: x in Solutions_of(A1,B1);
set AB=A1^^B1;
A37: len Segm(AB,SN,Seg width AB)=card SN by MATRIX_0:def 2;
A38: dom AB=Seg len AB & len AB=L by A20,FINSEQ_1:def 3,MATRIX_0:23;
reconsider x as Matrix of wA,wB,K by A20,A36,Th53;
A39: the_rank_of Segm(AB,Seg L,Seg width A1)=card N by A16,Th19;
A40: width AB=width A1+width B1 by A20,MATRIX_0:23;
now
let i such that
A41: i in SS;
A42: Line(A1,i)=width A1|->0.K by A28,A41;
A43: x in Solutions_of(A1,B1) & i in dom A1 & Line(A1,i)= width A1
|->0.K implies Line(B1,i)=width B1|->0.K by Th41;
thus Line(AB,i)=Line(A1,i)^Line(B1,i)by A35,A41,Th15
.=width AB|->0.K by A26,A35,A36,A40,A41,A42,A43,FINSEQ_2:123;
end;
then the_rank_of Segm(AB,SN,Seg width AB)=the_rank_of AB by A31,A38
,Th11;
then the_rank_of AB<=card SN by A37,MATRIX13:74;
then
A44: the_rank_of AB<=card N by FINSEQ_1:57;
width A1<=width AB by A40,NAT_1:11;
then Seg width A1 c= Seg width AB by FINSEQ_1:5;
then [:Seg L,Seg width A1:] c= Indices AB by A38,ZFMISC_1:95;
then card N <= the_rank_of AB by A39,MATRIX13:79;
then the_rank_of AB=card N by A44,XXREAL_0:1;
hence thesis by A16,A17;
end;
end;
suppose
A45: N={};
set ZERO=0.(K,L,wA);
A46: now
let i such that
A47: 1<=i & i<=L;
A48: dom A1=Seg len A1 & len A1=L by FINSEQ_1:def 3,MATRIX_0:def 2;
A49: i in Seg L by A47;
hence ZERO.i = wA|->0.K by FINSEQ_2:57
.= Line(A1,i) by A18,A45,A49,A48
.= A1.i by A49,MATRIX_0:52;
end;
A50: len ZERO=L by A20,MATRIX_0:23;
A51: width A1=wA by A20,MATRIX_0:23;
len A1=L by A20,MATRIX_0:23;
then ZERO=A1 by A50,A46;
then
A52: the_rank_of A=0 by A16,A50,A51,MATRIX13:95;
then
A53: ZERO=A by MATRIX13:95;
A54: Indices (A9^^B9)=[:Seg L,Seg (wA+wB):] by A20,MATRIX_0:23;
thus the_rank_of A=the_rank_of(A^^B) implies Solutions_of(A,B) is non
empty
proof
set x = the Matrix of wA,wB,K;
assume
A55: the_rank_of A = the_rank_of (A^^B);
Seg (wA+wB)\Seg wA c= Seg (wA+wB) by XBOOLE_1:36;
then
A56: [:Seg L,Seg (wA+wB)\Seg wA:] c= Indices (A^^B) by A54,ZFMISC_1:95;
Segm(A9^^B9,Seg L,Seg (wA+wB)\Seg wA)=B by Th19;
then 0=the_rank_of B by A52,A55,A56,MATRIX13:79;
then
A57: B=0.(K,L,wB) by A1,MATRIX13:95;
then wA=0 & wB=0 or Solutions_of(A9,B9) = the set of all
X where X is Matrix of wA,
wB,K by A2,A20,A53,Th54;
then Solutions_of(A9,B9)={{}} or x in Solutions_of(A9,B9) by A53,A57
,Th56;
hence thesis;
end;
A58: Indices B9=[:Seg L,Seg wB:] by A20,MATRIX_0:23;
A59: wA+wB = width (A9^^B9) by A20,MATRIX_0:23;
A60: Indices ZERO=[:Seg L,Seg wA:] by A20,MATRIX_0:23;
thus Solutions_of(A,B) is non empty implies the_rank_of A=the_rank_of(A
^^B)
proof
assume
A61: Solutions_of(A,B) is non empty;
assume the_rank_of A <> the_rank_of (A^^B);
then consider i,j such that
A62: [i,j] in Indices (A9^^B9) and
A63: (A9^^B9)*(i,j)<>0.K by A52,MATRIX13:94;
A64: j in Seg (wA+wB) by A59,A62,ZFMISC_1:87;
A65: dom Line(A9^^B9,i)=Seg (wA+wB) by A59,FINSEQ_2:124;
A66: len Line(A9,i)=wA by CARD_1:def 7;
A67: dom Line(B9,i)=Seg wB by FINSEQ_2:124;
A68: dom Line(A9,i)=Seg wA by FINSEQ_2:124;
A69: i in Seg L by A54,A62,ZFMISC_1:87;
then
A70: Line(A9^^B9,i)=Line(A9,i)^Line(B9,i) by Th15;
per cases by A64,A66,A65,A70,FINSEQ_1:25;
suppose
A71: j in dom Line(A9,i);
then
A72: [i,j] in Indices ZERO by A60,A69,A68,ZFMISC_1:87;
(A9^^B9)*(i,j) = Line(A9^^B9,i).j by A59,A64,MATRIX_0:def 7
.= Line(A9,i).j by A70,A71,FINSEQ_1:def 7
.= A9*(i,j) by A68,A71,MATRIX_0:def 7
.= 0.K by A53,A72,MATRIX_3:1;
hence thesis by A63;
end;
suppose
ex n st n in dom Line(B9,i) & j=wA + n;
then consider n such that
A73: n in dom Line(B9,i) and
A74: j=wA + n;
A75: [i,n] in Indices B by A58,A69,A67,A73,ZFMISC_1:87;
A76: B=0.(K,L,wB) by A53,A61,Th52;
(A9^^B9)*(i,j) = Line(A9^^B9,i).j by A59,A64,MATRIX_0:def 7
.= Line(B9,i).n by A66,A70,A73,A74,FINSEQ_1:def 7
.= B9*(i,n) by A67,A73,MATRIX_0:def 7
.= 0.K by A75,A76,MATRIX_3:1;
hence thesis by A63;
end;
end;
end;
end;
end;
begin :: Space of Solutions of Linear Equations
definition
let K;
let A be Matrix of K;
let b be FinSequence of K;
func Solutions_of(A,b) -> set equals
{f : ColVec2Mx f in Solutions_of(A,ColVec2Mx b
)};
coherence;
end;
theorem Th58:
for x st x in Solutions_of(A,ColVec2Mx b) ex f st x = ColVec2Mx
f & len f = width A
proof
let x such that
A1: x in Solutions_of(A,ColVec2Mx b);
consider X such that
A2: X=x and
A3: len X=width A and
A4: width X = width ColVec2Mx b and
A * X = ColVec2Mx b by A1;
per cases;
suppose
A5: len X=0;
take f=0|-> 0.K;
len (ColVec2Mx f)=0 by MATRIX_0:def 2;
hence thesis by A2,A3,A5,CARD_2:64;
end;
suppose
A6: len X>0;
take Col(X,1);
A7: len A=len (ColVec2Mx b) by A1,Th33;
len A<>0 by A3,A6,MATRIX_0:def 3;
then len b>0 by A7,MATRIX_0:def 2;
then width X=1 by A4,MATRIX_0:23;
hence thesis by A2,A3,A6,Th26,MATRIX_0:def 8;
end;
end;
theorem Th59:
for f st ColVec2Mx f in Solutions_of(A,ColVec2Mx b) holds len f = width A
proof
let f;
assume ColVec2Mx f in Solutions_of(A,ColVec2Mx b);
then
A1: ex g st ColVec2Mx f = ColVec2Mx g & len g = width A by Th58;
len ColVec2Mx f = len f by MATRIX_0:def 2;
hence thesis by A1,MATRIX_0:def 2;
end;
definition
let K;
let A be Matrix of K;
let b be FinSequence of K;
redefine func Solutions_of(A,b) -> Subset of (width A)-VectSp_over K;
coherence
proof
Solutions_of(A,b) c= the carrier of (width A)-VectSp_over K
proof
let x be object;
assume x in Solutions_of(A,b);
then consider f such that
A1: x=f and
A2: ColVec2Mx f in Solutions_of(A,ColVec2Mx b);
len f=width A by A2,Th59;
then (width A)-tuples_on the carrier of K= the carrier of (width A)
-VectSp_over K & f is Element of (width A)-tuples_on the carrier of K by
FINSEQ_2:92,MATRIX13:102;
hence thesis by A1;
end;
hence thesis;
end;
end;
registration
let K;
let A be Matrix of K;
let k be Element of NAT;
cluster Solutions_of(A,k|->0.K) -> linearly-closed for Subset of (width A)
-VectSp_over K;
coherence
proof
set V=(width A)-VectSp_over K;
set k0=k|->0.K;
set S=Solutions_of(A,k0);
A1: ColVec2Mx k0=0.(K,k,1) by Th32;
A2: now
let a being Element of K,v being Element of V;
assume v in S;
then consider f such that
A3: v=f and
A4: ColVec2Mx f in Solutions_of(A,ColVec2Mx k0);
reconsider f as Element of (width A)-tuples_on the carrier of K by A3,
MATRIX13:102;
now
per cases;
suppose
k=0;
then len (a*0.(K,k,1))=len 0.(K,k,1) & 0.(K,k,1)={} by MATRIX_3:def 5
;
hence 0.(K,k,1)=a*0.(K,k,1);
end;
suppose
A5: k>0;
then
A6: len 0.(K,k,1)=k & width 0.(K,k,1)=1 by MATRIX_0:23;
hence a*0.(K,k,1) = a*(0.K*0.(K,k,1)) by A5,MATRIX_5:24
.= (a*0.K)*0.(K,k,1) by MATRIX_5:11
.= 0.K*0.(K,k,1)
.= 0.(K,k,1) by A5,A6,MATRIX_5:24;
end;
end;
then a*ColVec2Mx f in Solutions_of(A,0.(K,k,1)) by A1,A4,Th35;
then ColVec2Mx (a*f) in Solutions_of(A,0.(K,k,1)) by Th30;
then a*f in Solutions_of(A,k0) by A1;
hence a*v in S by A3,MATRIX13:102;
end;
now
let v,u being Element of V such that
A7: v in S and
A8: u in S;
consider f such that
A9: v=f and
A10: ColVec2Mx f in Solutions_of(A,ColVec2Mx k0) by A7;
consider g such that
A11: u=g and
A12: ColVec2Mx g in Solutions_of(A,ColVec2Mx k0) by A8;
A13: len g=width A by A12,Th59;
reconsider f,g as Element of (width A)-tuples_on the carrier of K by A9
,A11,MATRIX13:102;
ColVec2Mx f+ColVec2Mx g in Solutions_of(A,0.(K,k,1)+0.(K,k,1)) by A1,A10
,A12,Th37;
then ColVec2Mx f+ColVec2Mx g in Solutions_of(A,0.(K,k,1)) by MATRIX_3:4;
then ColVec2Mx (f+g) in Solutions_of(A,0.(K,k,1))by A10,A13,Th28,Th59;
then f+g in S by A1;
hence v+u in S by A9,A11,MATRIX13:102;
end;
hence thesis by A2,VECTSP_4:def 1;
end;
end;
theorem Th60:
Solutions_of(A,b) is non empty & width A = 0 implies len A=0
proof
set S=Solutions_of(A,b);
assume that
A1: S is non empty and
A2: width A=0;
consider x being object such that
A3: x in S by A1;
consider f such that
x=f and
A4: ColVec2Mx f in Solutions_of(A,ColVec2Mx b) by A3;
consider X such that
ColVec2Mx f=X and
A5: len X = width A and
width X = width ColVec2Mx b and
A6: A * X = ColVec2Mx b by A4;
width (A * X) = width X by A5,MATRIX_3:def 4
.= 0 by A2,A5,MATRIX_0:def 3;
hence 0 = len b by A6,MATRIX_0:23
.= len ColVec2Mx b by MATRIX_0:def 2
.= len A by A4,Th33;
end;
theorem Th61:
width A <> 0 or len A = 0 implies Solutions_of(A,len A|->0.K) is non empty
proof
set L=len A|->0.K;
A1: len L=len A by CARD_1:def 7;
reconsider A9=A as Matrix of len A,width A,K by MATRIX_0:51;
assume
A2: width A <> 0 or len A = 0;
per cases by A2;
suppose
len A=0;
then Solutions_of(A9,ColVec2Mx L) = {{}} by A1,Th51;
then
A3: {} in Solutions_of(A9,ColVec2Mx L) by TARSKI:def 1;
then consider f such that
A4: {} = ColVec2Mx f and
len f = width A by Th58;
f in Solutions_of(A,L) by A3,A4;
hence thesis;
end;
suppose
A5: width A > 0;
ColVec2Mx L =0.(K,len A,1) by Th32;
then len ColVec2Mx L=len L & the_rank_of A=the_rank_of (A^^(ColVec2Mx L))
by Th23,MATRIX_0:def 2;
then Solutions_of(A,ColVec2Mx L) is non empty by A1,A5,Th57;
then consider x being object such that
A6: x in Solutions_of(A,ColVec2Mx L);
consider f such that
A7: x = ColVec2Mx f and
len f = width A by A6,Th58;
f in Solutions_of(A,L) by A6,A7;
hence thesis;
end;
end;
definition
let K;
let A be Matrix of K;
assume
A1: width A = 0 implies len A = 0;
func Space_of_Solutions_of(A) -> strict Subspace of (width A)-VectSp_over K
means
:Def5:
the carrier of it = Solutions_of(A,len A|->0.K);
existence
proof
Solutions_of(A,len A|->0.K) is non empty by A1,Th61;
hence thesis by VECTSP_4:34;
end;
uniqueness by VECTSP_4:29;
end;
theorem
for A be (Matrix of K), b be FinSequence of K st Solutions_of(A,b) is
non empty holds Solutions_of(A,b) is Coset of Space_of_Solutions_of A
proof
let A be (Matrix of K),b be FinSequence of K;
set V=(width A)-VectSp_over K;
reconsider B=b as Element of (len b)-tuples_on the carrier of K by
FINSEQ_2:92;
set CB=ColVec2Mx B;
assume Solutions_of(A,b) is non empty;
then consider x being object such that
A1: x in Solutions_of(A,B);
consider f such that
x=f and
A2: ColVec2Mx f in Solutions_of(A,CB) by A1;
set Cf=ColVec2Mx f;
A3: len f =width A by A2,Th59;
then reconsider f as Element of (width A)-tuples_on the carrier of K by
FINSEQ_2:92;
reconsider F=f as Element of V by MATRIX13:102;
A4: len CB=len B & len A=len CB by A2,Th33,MATRIX_0:def 2;
width A=0 implies len A=0 by A1,Th60;
then
A5: the carrier of Space_of_Solutions_of A=Solutions_of(A,len A|->0.K) by Def5;
A6: Solutions_of(A,b) c= F+ Space_of_Solutions_of A
proof
len B=len ((-1_K)*B) & (-1_K)*CB=ColVec2Mx ((-1_K)*B) by Th30,CARD_1:def 7;
then
A7: CB+(-1_K)*CB = ColVec2Mx (B+(-1_K)*B) by Th28
.= ColVec2Mx (B+-B) by FVSUM_1:59
.= ColVec2Mx (len A|->0.K) by A4,FVSUM_1:26;
let y be object;
assume y in Solutions_of(A,b);
then consider g such that
A8: y=g and
A9: ColVec2Mx g in Solutions_of(A,CB);
len g =width A by A9,Th59;
then reconsider g as Element of (width A)-tuples_on the carrier of K by
FINSEQ_2:92;
set Cg=ColVec2Mx g;
reconsider GF=g+(-1_K)*f as Element of V by MATRIX13:102;
A10: len f=len ((-1_K)*f) by A3,CARD_1:def 7;
width CB=width((-1_K)*CB) & (-1_K)*Cf in Solutions_of(A,(-1_K)*CB) by A2
,Th35,MATRIX_3:def 5;
then
A11: Cg+(-1_K)*Cf in Solutions_of(A,ColVec2Mx (len A|->0.K)) by A9,A7,Th37;
Cg+(-1_K)*Cf = Cg+ColVec2Mx ((-1_K)*f) by Th30
.=ColVec2Mx (g+(-1_K)*f) by A3,A9,A10,Th28,Th59;
then g+(-1_K)*f in Solutions_of(A,len A|->0.K) by A11;
then
A12: GF in Space_of_Solutions_of A by A5,STRUCT_0:def 5;
f+(g+(-1_K)*f) = f+((-1_K)*f+g) by FINSEQOP:33
.= f+(-f+g) by FVSUM_1:59
.= (f+-f)+g by FINSEQOP:28
.= (width A|->0.K)+g by FVSUM_1:26
.= g by FVSUM_1:21;
then F+GF=g by MATRIX13:102;
hence thesis by A8,A12;
end;
F+ Space_of_Solutions_of A c=Solutions_of(A,b)
proof
let y be object;
A13: len A=0 or len A<>0;
len (len A|-> 0.K) =len A by CARD_1:def 7;
then
A14: width CB=1 & width ColVec2Mx (len A|->0.K)=1 or width CB=0 & width
ColVec2Mx (len A|->0.K)=0 by A4,A13,Th26,MATRIX_0:22;
ColVec2Mx (len A|->0.K)=0.(K,len A,1) by Th32;
then
A15: CB+ColVec2Mx (len A|->0.K)=CB by A4,MATRIX_3:4;
assume y in F+ Space_of_Solutions_of A;
then consider U be Element of V such that
A16: U in Space_of_Solutions_of A and
A17: y=F+U by VECTSP_4:42;
reconsider u=U as Element of (width A)-tuples_on the carrier of K by
MATRIX13:102;
u in Solutions_of(A,len A|->0.K) by A5,A16,STRUCT_0:def 5;
then consider g such that
A18: u=g and
A19: ColVec2Mx g in Solutions_of(A,ColVec2Mx (len A|->0.K));
width A = len g by A19,Th59;
then ColVec2Mx (f+g)=Cf+ColVec2Mx g by A2,Th28,Th59;
then ColVec2Mx (f+g) in Solutions_of(A,CB) by A2,A19,A14,A15,Th37;
then f+g in Solutions_of(A,b);
hence thesis by A17,A18,MATRIX13:102;
end;
then F+ Space_of_Solutions_of A =Solutions_of(A,b) by A6;
hence thesis by VECTSP_4:def 6;
end;
theorem Th63:
for A st ( width A = 0 implies len A = 0 ) & the_rank_of A = 0
holds Space_of_Solutions_of A = (width A)-VectSp_over K
proof
let A such that
A1: width A = 0 implies len A = 0 and
A2: the_rank_of A = 0;
set L=len A|->0.K;
the carrier of (width A)-VectSp_over K c= Solutions_of(A,L)
proof
let x be object such that
A3: x in the carrier of (width A)-VectSp_over K;
reconsider x9=x as Element of (width A)-tuples_on the carrier of K by A3,
MATRIX13:102;
A4: A=0.(K,len A,width A) & ColVec2Mx L=0.(K,len A,1) by A2,Th32,MATRIX13:95;
per cases;
suppose
A5: len A=0;
then Solutions_of(A,ColVec2Mx L)={{}} by A4,Th51;
then
A6: {} in Solutions_of(A,ColVec2Mx L) by TARSKI:def 1;
then consider f such that
A7: {} = ColVec2Mx f and
A8: len f = width A by Th58;
width A=0 by A5,MATRIX_0:def 3;
then
the carrier of (width A)-VectSp_over K = 0-tuples_on the carrier of
K by MATRIX13:102
.= {<*>the carrier of K} by FINSEQ_2:94;
then x=<*>the carrier of K by A3,TARSKI:def 1;
then f=x by A5,A8,MATRIX_0:def 3;
hence thesis by A6,A7;
end;
suppose
A9: len A>0;
A10: len x9=width A by CARD_1:def 7;
Solutions_of(A,ColVec2Mx L)= the set of all
X where X is Matrix of width A,1,K by A1,A4,A9,Th54;
then ColVec2Mx x9 in Solutions_of(A,ColVec2Mx L) by A10;
hence thesis;
end;
end;
then the carrier of (width A)-VectSp_over K = Solutions_of(A,L)
.= the carrier of Space_of_Solutions_of A by A1,Def5;
hence thesis by VECTSP_4:31;
end;
theorem
for A st Space_of_Solutions_of A = (width A)-VectSp_over K holds
the_rank_of A = 0
proof
let A such that
A1: Space_of_Solutions_of A = (width A)-VectSp_over K;
assume the_rank_of A <> 0;
then consider i,j such that
A2: [i,j] in Indices A and
A3: A*(i,j) <> 0.K by MATRIX13:94;
A4: j in Seg width A by A2,ZFMISC_1:87;
then A5: width A <> 0;
set L=Line(1.(K,width A),j);
A6: width 1.(K,width A)=width A by MATRIX_0:24;
then
A7: dom L=Seg width A by FINSEQ_2:124;
A8: Indices 1.(K,width A)=[:Seg width A,Seg width A:] by MATRIX_0:24;
A9: now
let k such that
A10: k in dom L and
A11: k<>j;
[j,k] in Indices 1.(K,width A) by A4,A7,A8,A10,ZFMISC_1:87;
hence 0.K = 1.(K,width A)*(j,k) by A11,MATRIX_1:def 3
.= L.k by A6,A7,A10,MATRIX_0:def 7;
end;
A12: dom Line(A,i)=Seg width A by FINSEQ_2:124;
[j,j] in Indices 1.(K,width A) by A4,A8,ZFMISC_1:87;
then 1_K = 1.(K,width A)*(j,j) by MATRIX_1:def 3
.= L.j by A4,A6,MATRIX_0:def 7;
then
A13: Sum(mlt(L,Line(A,i))) = Line(A,i).j by A4,A7,A12,A9,MATRIX_3:17
.= A*(i,j) by A4,MATRIX_0:def 7;
A14: ColVec2Mx (len A|->0.K)=0.(K,len A,1) by Th32;
A15: i in dom A by A2,ZFMISC_1:87;
L in (width A)-tuples_on the carrier of K by A6;
then L in the carrier of Space_of_Solutions_of A by A1,MATRIX13:102;
then L in Solutions_of(A,len A|->0.K) by Def5,A5;
then consider f such that
A16: f=L and
A17: ColVec2Mx f in Solutions_of(A,ColVec2Mx (len A|->0.K));
consider X such that
A18: X=ColVec2Mx f and
A19: len X = width A and
width X = width ColVec2Mx (len A|->0.K) and
A20: A * X = ColVec2Mx (len A|->0.K) by A17;
A21: 1 in Seg 1;
A22: dom A=Seg len A by FINSEQ_1:def 3;
then len A<>0 by A2,ZFMISC_1:87;
then Indices ColVec2Mx (len A|->0.K)=[:Seg len A,Seg 1:] by A14,MATRIX_0:23;
then
A23: [i,1] in Indices ColVec2Mx (len A|->0.K) by A15,A22,A21,ZFMISC_1:87;
then Line(A,i)"*"Col(X,1) = 0.(K,len A,1)*(i,1) by A19,A20,A14,MATRIX_3:def 4
.= 0.K by A14,A23,MATRIX_3:1;
then 0.K = Col(X,1)"*"Line(A,i) by FVSUM_1:90
.= Sum(mlt(f,Line(A,i))) by A18,A19,Th26,A5;
hence thesis by A3,A16,A13;
end;
theorem Th65:
for i,j st j in Seg m & n>0 & (i = j implies a <> -1_K) holds
Space_of_Solutions_of A9 = Space_of_Solutions_of RLine(A9,i,Line(A9,i) + a*Line
(A9,j))
proof
let i,j such that
A1: j in Seg m and
A2: n>0 and
A3: i = j implies a <> -1_K;
set L=len A9|->0.K;
set R=RLine(A9,i,Line(A9,i) + a*Line(A9,j));
A4: m<>0 by A1;
then
A5: width R=n by MATRIX_0:23;
len L=len A9 by CARD_1:def 7;
then len L = m by MATRIX_0:def 2;
then reconsider C=ColVec2Mx L as Matrix of m,1,K;
set RC=RLine(C,i,Line(C,i) + a*Line(C,j));
A6: C=0.(K,len A9,1) by Th32;
now
let i9,j9 be Nat such that
A7: [i9,j9] in Indices C;
reconsider I=i9,J=j9 as Element of NAT by ORDINAL1:def 12;
A8: len (Line(C,i) + a*Line(C,j))=width C by CARD_1:def 7;
now
per cases;
suppose
A9: i9=i;
A10: 1=width C by A4,MATRIX_0:23;
then
A11: j9 in Seg 1 by A7,ZFMISC_1:87;
then Line(C,j).j9=C*(j,j9) by A10,MATRIX_0:def 7;
then
A12: (a*Line(C,j)).j9=a*(C*(j,j9)) by A10,A11,FVSUM_1:51;
Indices C=[:Seg m,Seg 1:] by A4,MATRIX_0:23;
then
A13: [j,j9] in Indices C by A1,A11,ZFMISC_1:87;
Line(C,i).j9=C*(i,j9) by A10,A11,MATRIX_0:def 7;
then (Line(C,i) + a*Line(C,j)).j9 = C*(i,j9)+a*(C*(j,j9)) by A10,A11
,A12,FVSUM_1:18
.= 0.K +a*(C*(j,j9)) by A6,A7,A9,MATRIX_3:1
.= 0.K +a*0.K by A6,A13,MATRIX_3:1
.= 0.K +0.K
.= 0.K by RLVECT_1:def 4
.= C*(i9,j9) by A6,A7,MATRIX_3:1;
hence C*(I,J)=RC*(I,J) by A7,A8,A9,MATRIX11:def 3;
end;
suppose
i<>i9;
hence C*(I,J)=RC*(I,J) by A7,A8,MATRIX11:def 3;
end;
end;
hence C*(i9,j9)=RC*(i9,j9);
end;
then RC=C by MATRIX_0:27;
then
A14: Solutions_of(A9,C)=Solutions_of(R,C) by A1,A3,Th40;
set SR=Space_of_Solutions_of R;
len A9=m & len R=m by A4,MATRIX_0:23;
then
A15: the carrier of SR = Solutions_of(R,L) by A2,A5,Def5;
set SA=Space_of_Solutions_of A9;
A16: width A9=n by A4,MATRIX_0:23;
then the carrier of SA = Solutions_of(A9,L) by A2,Def5;
hence thesis by A16,A5,A14,A15,VECTSP_4:29;
end;
theorem Th66:
for N st N c= dom A & N is non empty & width A > 0 & for i st i
in (dom A) \ N holds Line(A,i) = width A |-> 0.K holds Space_of_Solutions_of A
= Space_of_Solutions_of Segm(A,N,Seg width A)
proof
let N such that
A1: N c= dom A and
A2: N is non empty and
A3: width A>0 and
A4: for i st i in (dom A) \ N holds Line(A,i) = width A |-> 0.K;
set L=len A|->0.K;
set C=ColVec2Mx L;
A5: len L=len A by CARD_1:def 7;
set S=Segm(A,N,Seg width A);
A6: width S=card Seg width A by A2,MATRIX_0:23;
then
A7: width A=width S by FINSEQ_1:57;
set SS=Space_of_Solutions_of S;
len S=card N by MATRIX_0:def 2;
then
A8: the carrier of SS = Solutions_of(S,card N|->0.K) by A3,A6,Def5;
set SA=Space_of_Solutions_of A;
A9: the carrier of SA = Solutions_of(A,L) by A3,Def5;
A10: C=0.(K,len A,1) by Th32;
len C=len L by MATRIX_0:def 2;
then
A11: dom C=dom A by A5,FINSEQ_3:29;
A12: dom A=Seg len A by FINSEQ_1:def 3;
then
A13: Seg len A<>{} by A1,A2;
then
A14: width C=1 by Th26;
then
A15: card Seg width C=1 by FINSEQ_1:57;
now
A16: rng Sgm Seg 1=Seg 1 by FINSEQ_1:def 13;
A17: Indices Segm(C,N,Seg width C)=Indices 0.(K,card N,1) by A15,MATRIX_0:26;
let k,l such that
A18: [k,l] in Indices Segm(C,N,Seg width C);
reconsider kk=k,ll=l as Element of NAT by ORDINAL1:def 12;
[:N,Seg width C:] c= Indices C & rng Sgm N=N by A1,A12,A11,FINSEQ_1:def 13
,ZFMISC_1:95;
then
A19: [Sgm N.kk,Sgm Seg width C.ll] in Indices C by A14,A18,A16,MATRIX13:17;
thus Segm(C,N,Seg width C)*(k,l) = C*(Sgm N.kk,Sgm Seg width C.ll) by A18,
MATRIX13:def 1
.= 0.K by A10,A19,MATRIX_3:1
.=0.(K,card N,1)*(k,l) by A18,A17,MATRIX_3:1;
end;
then
A20: Segm(C,N,Seg width C) = 0.(K,card N,1) by A15,MATRIX_0:27
.= ColVec2Mx (card N|->0.K) by Th32;
now
let i such that
A21: i in (dom A) \ N;
A22: i in dom A by A21,XBOOLE_0:def 5;
then Line(C,i) = C.i by A5,A12,MATRIX_0:52
.= (len A|->(width C|->0.K)).i by A10,A13,Th26
.= width C|->0.K by A12,A22,FINSEQ_2:57;
hence Line(A,i)=width A|->0.K & Line(C,i)=width C|->0.K by A4,A21;
end;
then
Solutions_of(A,C)=Solutions_of(Segm(A,N,Seg width A),Segm(C,N,Seg width
C)) by A1,A2,A11,Th45;
hence thesis by A20,A7,A9,A8,VECTSP_4:29;
end;
Lm7: for A be set st A c= dom g ex ga,gb be FinSequence of K st ga=g*Sgm A &
gb=g*Sgm (dom g\A) & Sum g = Sum ga + Sum gb
proof
A1: rng g c= the carrier of K by FINSEQ_1:def 4;
set Ad=the addF of K;
A2: dom g=Seg len g by FINSEQ_1:def 3;
then
A3: dom g=rng idseq len g & dom g=dom idseq len g;
let A be set such that
A4: A c= dom g;
A5: rng Sgm(A)=A by A4,A2,FINSEQ_1:def 13;
A6: (idseq len g)"A=A by A4,A2,FUNCT_2:94;
A7: dom g\A c= dom g by XBOOLE_1:36;
then
A8: rng Sgm(dom g\A)=dom g\A by A2,FINSEQ_1:def 13;
then reconsider ga=g*Sgm A,gb=g*Sgm (dom g\A) as FinSequence by A4,A5,
FINSEQ_1:16,XBOOLE_1:36;
(idseq len g)"(dom g\A)=dom g\A by A2,A7,FUNCT_2:94;
then
A9: Sgm(A)^Sgm(dom g\A) is Permutation of dom g by A6,A3,FINSEQ_3:114;
then reconsider gS=g*(Sgm(A)^Sgm(dom g\A)) as FinSequence of K by A2,
FINSEQ_2:46;
rng ga c=rng g by RELAT_1:26;
then
A10: rng ga c= the carrier of K by A1;
rng gb c=rng g by RELAT_1:26;
then rng gb c= the carrier of K by A1;
then reconsider ga,gb as FinSequence of K by A10,FINSEQ_1:def 4;
take ga,gb;
Ad $$ g = Ad "**" gS by A9,FINSOP_1:7,FVSUM_1:8
.= Ad"**"(ga^gb) by A4,A7,A5,A8,Th5
.= Sum ga+Sum gb by FINSOP_1:5,FVSUM_1:8;
hence thesis;
end;
theorem Th67:
for A be Matrix of n,m,K, N st card N = n & N c= Seg m & Segm(A,
Seg n,N) = 1.(K,n) & n > 0 & m-'n>0 ex MVectors be Matrix of m-'n,m,K st Segm(
MVectors,Seg (m-'n),Seg m \ N) = 1.(K,m-'n) & Segm(MVectors,Seg (m-'n),N) = -
Segm(A,Seg n,Seg m \N)@ & Lin(lines MVectors) = Space_of_Solutions_of A
proof
let A be Matrix of n,m,K, N such that
A1: card N = n and
A2: N c= Seg m and
A3: Segm(A,Seg n,N) = 1.(K,n) and
A4: n > 0 and
A5: m-'n>0;
Seg m <>{} by A1,A2,A4,CARD_1:27,XBOOLE_1:3;
then
A6: m<>0;
consider MV be Matrix of m-'n,m,K such that
A7: Segm(MV,Seg (m-'n),Seg m \ N) = 1.(K,m-'n) and
A8: Segm(MV,Seg (m-'n),N) = -Segm(A,Seg n,Seg m \N)@ and
A9: for l for M be Matrix of m,l,K st for i st i in Seg l holds (ex j
st j in Seg (m-'n) & Col(M,i) = Line(MV,j)) or Col(M,i) = m|->0.K holds M in
Solutions_of(A,0.(K,n,l)) by A1,A2,A3,A4,Th49;
A10: width MV=m by A5,MATRIX_0:23;
A11: len MV=m-'n & len MV >=the_rank_of MV by MATRIX13:74,MATRIX_0:def 2;
A12: Indices MV=[:Seg (m-'n),Seg m:] by A5,MATRIX_0:23;
Seg m \ N c= Seg m by XBOOLE_1:36;
then
A13: [:Seg (m-'n),Seg m \ N:] c= Indices MV by A12,ZFMISC_1:95;
A14: width A=m by A4,MATRIX_0:23;
A15: len A=n by A4,MATRIX_0:23;
lines MV c= Solutions_of(A,len A|->0.K)
proof
let x be object;
assume x in lines MV;
then consider k such that
A16: k in Seg (m-'n) and
A17: x = Line(MV,k) by MATRIX13:103;
set C=ColVec2Mx Line(MV,k);
A18: m = width MV by A5,MATRIX_0:23
.= len Line(MV,k) by CARD_1:def 7;
now
let i;
assume i in Seg 1;
then
A19: i=1 by FINSEQ_1:2,TARSKI:def 1;
Col(C,1)=Line(MV,k) by A6,A18,Th26;
hence (ex j st j in Seg (m-'n) & Col(C,i) = Line(MV,j)) or Col(C,i) = m
|->0.K by A16,A19;
end;
then C in Solutions_of(A,0.(K,n,1)) by A9,A18;
then C in Solutions_of(A,ColVec2Mx(len A|->0.K))by A15,Th32;
hence thesis by A17;
end;
then Lin (lines MV) is Subspace of Lin (Solutions_of(A,len A|->0.K)) by A14,
VECTSP_7:13;
then
A20: the carrier of Lin (lines MV) c= the carrier of Lin (Solutions_of(A,len
A|->0.K)) by VECTSP_4:def 2;
m-'n+0>0 by A5;
then m-'n>=1 by NAT_1:19;
then
A21: Det 1.(K,m-'n)=1_K by MATRIX_7:16;
A22: 0.K<>1_K;
A23: card Seg m=m by FINSEQ_1:57;
then
A24: m-'n=m-n by A1,A2,NAT_1:43,XREAL_1:233;
A25: card (Seg m \ N) =m-n by A1,A2,A23,CARD_2:44;
then
A26: card Seg (m-'n)=card (Seg m \ N) by A24,FINSEQ_1:57;
EqSegm(MV,Seg (m-'n),Seg m \ N)=1.(K,m-'n) by A7,A24,A25,FINSEQ_1:57
,MATRIX13:def 3;
then m-'n <= the_rank_of MV by A24,A25,A26,A13,A21,A22,MATRIX13:def 4;
then
A27: the_rank_of MV = m-'n by A11,XXREAL_0:1;
A28: the carrier of Space_of_Solutions_of A c= the carrier of Lin(lines MV)
proof
set SN=Seg m\N;
let y be object;
assume y in the carrier of Space_of_Solutions_of A;
then y in Solutions_of(A,len A|->0.K) by A6,A14,Def5;
then consider f such that
A29: f=y and
A30: ColVec2Mx f in Solutions_of(A,ColVec2Mx (len A|->0.K));
A31: len f=m by A14,A30,Th59;
deffunc F(Nat)=f/.(Sgm SN.$1)*Line(MV,$1);
A32: dom f=Seg len f by FINSEQ_1:def 3;
consider M be FinSequence of (width MV)-tuples_on the carrier of K such
that
A33: len M= m-'n and
A34: for j st j in dom M holds M.j = F(j) from FINSEQ_2:sch 1;
A35: dom M = Seg(m-'n) by A33,FINSEQ_1:def 3;
reconsider M as FinSequence of the carrier of m-VectSp_over K by A10,
MATRIX13:102;
reconsider M1=FinS2MX M as Matrix of m-'n,m,K by A33;
now
let i such that
A36: i in Seg (m-'n);
take a=f/.(Sgm SN.i);
thus Line(M1,i) = M1.i by A36,MATRIX_0:52
.= a * Line(MV,i) by A34,A35,A36;
end;
then consider L be Linear_Combination of lines MV such that
A37: L (#) MX2FinS MV = M1 by A27,MATRIX13:105,108;
reconsider SumL=Sum L,f as Element of m-tuples_on the carrier of K by A31,
FINSEQ_2:92,MATRIX13:102;
now
let i such that
A38: i in Seg m;
A39: SN c= Seg m by XBOOLE_1:36;
A40: now
per cases;
suppose
A41: i in N;
consider X such that
A42: X = ColVec2Mx f and
A43: len X = width A and
width X=width ColVec2Mx(len A|->0.K) and
A44: A * X = ColVec2Mx (len A|->0.K) by A30;
A45: f = Col(X,1) by A6,A42,Th26;
reconsider F=f as Element of (width A)-tuples_on the carrier of K by
A4,MATRIX_0:23;
A46: rng Sgm Seg (m-'n)=Seg (m-'n) by FINSEQ_1:def 13;
A47: rng Sgm N=N by A2,FINSEQ_1:def 13;
then consider x being object such that
A48: x in dom Sgm N and
A49: Sgm N.x=i by A41,FUNCT_1:def 3;
reconsider x as Element of NAT by A48;
set L=Line(A,x);
A50: dom mlt(L,F)=Seg m by A14,FINSEQ_2:124;
then consider mN,mSN be FinSequence of K such that
A51: mN = mlt(L,F)*Sgm N and
A52: mSN = mlt(L,F)*Sgm SN and
A53: Sum mlt(L,F) = Sum mN + Sum mSN by A2,Lm7;
A54: dom Sgm N=Seg card N by A2,FINSEQ_3:40;
then
A55: dom mN=Seg n by A1,A2,A47,A50,A51,RELAT_1:27;
Indices 1.(K,n)=[:Seg n,Seg n:] by MATRIX_0:24;
then
A56: [x,x] in Indices 1.(K,n) by A1,A54,A48,ZFMISC_1:87;
A57: Sgm N.x in N by A47,A48,FUNCT_1:def 3;
then
A58: F.(Sgm N.x)=F/.(Sgm N.x) & L.(Sgm N.x)=A*(x,Sgm N. x) by A2,A14,A31
,A32,MATRIX_0:def 7,PARTFUN1:def 6;
A59: x = (idseq n).x by A1,A54,A48,FINSEQ_2:49
.= Sgm Seg n.x by FINSEQ_3:48;
now
let j such that
A60: j in dom mN and
A61: j<>x;
Indices 1.(K,n)=[:Seg n,Seg n:] by MATRIX_0:24;
then
A62: [x,j] in Indices 1.(K,n) by A1,A54,A48,A55,A60,ZFMISC_1:87;
A63: Sgm N.j in N by A1,A47,A54,A55,A60,FUNCT_1:def 3;
then
A64: F.(Sgm N.j)=F/.(Sgm N.j) & L.(Sgm N.j)=A*(x,Sgm N. j) by A2,A14,A31
,A32,MATRIX_0:def 7,PARTFUN1:def 6;
thus mN.j = mlt(L,F).(Sgm N.j) by A51,A60,FUNCT_1:12
.= F/.(Sgm N.j)*(A*(Sgm Seg n.x,Sgm N.j)) by A2,A14,A59,A63,A64,
FVSUM_1:61
.= F/.(Sgm N.j)*(1.(K,n)*(x,j)) by A3,A62,MATRIX13:def 1
.= F/.(Sgm N.j)*0.K by A61,A62,MATRIX_1:def 3
.= 0.K;
end;
then Sum mN=mN.x by A1,A54,A48,A55,MATRIX_3:12;
then
A65: Sum mN = mlt(L,F).(Sgm N.x) by A1,A54,A48,A51,A55,FUNCT_1:12
.= F/.(Sgm N.x)*(A*(Sgm Seg n.x,Sgm N.x)) by A2,A14,A59,A57,A58,
FVSUM_1:61
.= F/.(Sgm N.x)*(1.(K,n)*(x,x)) by A3,A56,MATRIX13:def 1
.= F/.(Sgm N.x)*1_K by A56,MATRIX_1:def 3
.= F/.i by A49,VECTSP_1:def 4
.= f.i by A31,A32,A38,PARTFUN1:def 6;
A66: dom Sgm SN=Seg card SN by FINSEQ_3:40,XBOOLE_1:36;
A67: rng Sgm SN=SN by A39,FINSEQ_1:def 13;
then dom mSN=Seg (m-'n) by A24,A25,A50,A52,A66,RELAT_1:27,XBOOLE_1:36
;
then
A68: len mSN=m-'n by FINSEQ_1:def 3;
A69: ColVec2Mx (len A|->0.K)=0.(K,len A,1) by Th32;
Indices 0.(K,len A,1)=[:Seg len A,Seg 1:] & 1 in Seg 1 by A4,A15,
MATRIX_0:23;
then
A70: [x,1] in Indices 0.(K,len A,1)by A1,A15,A54,A48,ZFMISC_1:87;
then
A71: 0.K = (ColVec2Mx (len A|->0.K))*(x,1) by A69,MATRIX_3:1
.= Line(A,x)"*"Col(X,1) by A43,A44,A69,A70,MATRIX_3:def 4
.= Sum(mlt(Line(A,x),Col(X,1)));
reconsider mSN as Element of (m-'n)-tuples_on the carrier of K by A68
,FINSEQ_2:92;
A72: width M1=m by A5,MATRIX_0:23;
now
let j such that
A73: j in Seg (m-'n);
A74: j = idseq(m-'n).j by A73,FINSEQ_2:49
.= Sgm Seg (m-'n).j by FINSEQ_3:48;
A75: Line(M1,j) = M1.j by A73,MATRIX_0:52
.= f/.(Sgm SN.j)*Line(MV,j) by A34,A35,A73;
A76: Sgm SN.j in SN by A24,A25,A67,A66,A73,FUNCT_1:def 3;
then f/.(Sgm SN.j) = F.(Sgm SN.j) & A*(x,Sgm SN.j) = L.(Sgm SN.j)
by A14,A31,A32,A39,MATRIX_0:def 7,PARTFUN1:def 6;
then
A77: f/.(Sgm SN.j) * (A*(x,Sgm SN.j)) = mlt(L,F).(Sgm SN.j) by A14,A39
,A76,FVSUM_1:61
.= mSN.j by A24,A25,A52,A66,A73,FUNCT_1:13;
A78: x = idseq(n).x by A1,A54,A48,FINSEQ_2:49
.= Sgm Seg n.x by FINSEQ_3:48;
A79: [:Seg (m-'n),N:] c= Indices MV by A2,A12,ZFMISC_1:95;
[j,i] in Indices MV by A2,A12,A41,A73,ZFMISC_1:87;
then
A80: [j,x] in Indices Segm(MV,Seg (m-'n),N) by A47,A49,A46,A74,A79,
MATRIX13:17;
then
A81: [j,x] in Indices (Segm(A,Seg n,SN)@) by A8,Lm1;
then
A82: [x,j] in Indices Segm(A,Seg n,SN) by MATRIX_0:def 6;
A83: Line(MV,j).i = MV*(Sgm Seg (m-'n).j,Sgm N.x) by A10,A38,A49,A74,
MATRIX_0:def 7
.= (-Segm(A,Seg n,Seg m \N)@)*(j,x) by A8,A80,MATRIX13:def 1
.= -((Segm(A,Seg n,Seg m \N)@)*(j,x)) by A81,MATRIX_3:def 2
.= -(Segm(A,Seg n,SN)*(x,j)) by A82,MATRIX_0:def 6
.= -(A*(x,Sgm SN.j)) by A82,A78,MATRIX13:def 1;
dom M1=Seg (m-'n) by A33,FINSEQ_1:def 3;
hence Col(M1,i).j = M1*(j,i) by A73,MATRIX_0:def 8
.= Line(M1,j).i by A38,A72,MATRIX_0:def 7
.= f/.(Sgm SN.j) * (-(A*(x,Sgm SN.j))) by A2,A10,A41,A75,A83,
FVSUM_1:51
.= -(f/.(Sgm SN.j) * (A*(x,Sgm SN.j))) by VECTSP_1:8
.= (-mSN).j by A73,A77,FVSUM_1:23;
end;
then Col(M1,i)=-mSN by A33,FINSEQ_2:119;
hence Sum Col(M1,i) = - Sum mSN by FVSUM_1:75
.= -Sum mSN +(Sum mSN+Sum mN) by A71,A45,A53,RLVECT_1:def 4
.= (-Sum mSN+Sum mSN)+Sum mN by RLVECT_1:def 3
.= 0.K+Sum mN by VECTSP_1:19
.= f.i by A65,RLVECT_1:def 4;
end;
suppose
A84: not i in N;
A85: rng Sgm SN=SN by A39,FINSEQ_1:def 13;
i in SN by A38,A84,XBOOLE_0:def 5;
then consider x being object such that
A86: x in dom Sgm SN and
A87: Sgm SN.x=i by A85,FUNCT_1:def 3;
reconsider x as Element of NAT by A86;
A88: dom Sgm SN=Seg card SN by FINSEQ_3:40,XBOOLE_1:36;
then
A89: Line(M1,x) = M1.x by A24,A25,A86,MATRIX_0:52
.= f/.(Sgm SN.x)*Line(MV,x) by A24,A25,A34,A35,A88,A86;
[x,x] in [:Seg (m-'n),Seg (m-'n):] by A24,A25,A88,A86,ZFMISC_1:87;
then
A90: [x,x] in Indices 1.(K,m-'n) by MATRIX_0:24;
x = idseq (m-'n).x by A24,A25,A88,A86,FINSEQ_2:49
.= Sgm Seg (m-'n).x by FINSEQ_3:48;
then
A91: Line(MV,x).i = MV*(Sgm Seg (m-'n).x,Sgm SN.x) by A10,A38,A87,
MATRIX_0:def 7
.= 1.(K,m-'n)*(x,x) by A7,A90,MATRIX13:def 1
.= 1_K by A90,MATRIX_1:def 3;
A92: dom Col(M1,i)=Seg len M1 by FINSEQ_2:124;
A93: dom M1=Seg len M1 by FINSEQ_1:def 3;
A94: len M1=m-n by A5,A24,MATRIX_0:23;
A95: width M1=m by A5,MATRIX_0:23;
A96: now
let j such that
A97: j in dom Col(M1,i) and
A98: x<>j;
A99: Line(M1,j) = M1.j by A92,A97,MATRIX_0:52
.= f/.(Sgm SN.j)*Line(MV,j) by A34,A92,A93,A97;
[j,x] in [:Seg (m-'n),Seg (m-'n):] by A24,A25,A88,A86,A92,A94,A97,
ZFMISC_1:87;
then
A100: [j,x] in Indices 1.(K,m-'n) by MATRIX_0:24;
j = idseq (m-'n).j by A24,A92,A94,A97,FINSEQ_2:49
.= Sgm Seg (m-'n).j by FINSEQ_3:48;
then
A101: Line(MV,j).i = MV*(Sgm Seg (m-'n).j,Sgm SN.x) by A10,A38,A87,
MATRIX_0:def 7
.= 1.(K,m-'n)*(j,x) by A7,A100,MATRIX13:def 1
.= 0.K by A98,A100,MATRIX_1:def 3;
thus Col(M1,i).j = M1*(j,i) by A92,A93,A97,MATRIX_0:def 8
.= (f/.(Sgm SN.j)*Line(MV,j)).i by A38,A95,A99,MATRIX_0:def 7
.= (f/.(Sgm SN.j))*0.K by A10,A38,A101,FVSUM_1:51
.= 0.K;
end;
Col(M1,i).x = M1*(x,i) by A25,A88,A86,A94,A93,MATRIX_0:def 8
.= Line(M1,x).i by A38,A95,MATRIX_0:def 7
.= (f/.(Sgm SN.x))*1_K by A10,A38,A91,A89,FVSUM_1:51
.= (f/.i) by A87,VECTSP_1:def 4
.= f.i by A31,A32,A38,PARTFUN1:def 6;
hence Sum Col(M1,i)=f.i by A25,A88,A86,A92,A94,A96,MATRIX_3:12;
end;
end;
Carrier L c= lines MV by VECTSP_6:def 4;
hence SumL.i=f.i by A27,A37,A38,A40,MATRIX13:105,107;
end;
then
A102: SumL=f by FINSEQ_2:119;
the carrier of Lin(lines MV) = the set of all
Sum(l) where l is Linear_Combination
of lines MV by VECTSP_7:def 2;
hence thesis by A29,A102;
end;
take MV;
Solutions_of(A,len A|->0.K)=the carrier of Space_of_Solutions_of A by A6,A14
,Def5;
then
the carrier of Lin (lines MV) c= the carrier of Space_of_Solutions_of(A
) by A20,VECTSP_7:11;
then the carrier of Lin (lines MV) = the carrier of Space_of_Solutions_of(A
) by A28;
hence thesis by A14,A7,A8,VECTSP_4:29;
end;
Lm8: dim Space_of_Solutions_of 1.(K,n) = 0
proof
set ONE=1.(K,n);
set SS=Space_of_Solutions_of ONE;
A1: the carrier of SS c= the carrier of (0).SS
proof
let x be object such that
A2: x in the carrier of SS;
A3: len ONE=n by MATRIX_0:24;
A4: width ONE=n by MATRIX_0:24;
then width ONE=0 implies len ONE=0 by MATRIX_0:24;
then x in Solutions_of(ONE,n|->0.K) by A2,A3,Def5;
then consider f such that
A5: f=x and
A6: ColVec2Mx f in Solutions_of(ONE,ColVec2Mx (n|->0.K));
consider X such that
A7: X =ColVec2Mx f and
A8: len X = width ONE and
width X = width ColVec2Mx (n|->0.K) and
A9: ONE * X = ColVec2Mx (n|->0.K) by A6;
A10: now
per cases;
suppose
A11: n>0;
ONE*X=X by A4,A8,MATRIXR2:68;
hence n|->0.K = Col(X,1) by A9,A11,Th26
.= f by A4,A7,A8,A11,Th26;
end;
suppose
A12: n=0;
then f={} by A4,A7,A8,MATRIX_0:def 2;
hence f=n|->0.K by A12;
end;
end;
0.SS = 0.((width ONE)-VectSp_over K ) by VECTSP_4:11
.= n|->0.K by A4,MATRIX13:102;
then f in {0.SS} by A10,TARSKI:def 1;
hence thesis by A5,VECTSP_4:def 3;
end;
the carrier of (0).SS c= the carrier of SS by VECTSP_4:def 2;
then the carrier of SS = the carrier of (0).SS by A1;
then (0).SS = (Omega).SS by VECTSP_4:29;
hence thesis by VECTSP_9:29;
end;
:: The dimension of Space of Solutions of linear equations
theorem Th68:
for A st ( width A = 0 implies len A = 0 ) holds dim
Space_of_Solutions_of A = width A - the_rank_of A
proof
let A such that
A1: width A = 0 implies len A = 0;
set W=width A;
set L=len A;
reconsider A9=A as Matrix of L,W,K by MATRIX_0:51;
per cases;
suppose
A2: the_rank_of A=0;
dim ((width A)-VectSp_over K)=width A by MATRIX13:112;
hence thesis by A1,A2,Th63;
end;
suppose
A3: the_rank_of A>0;
defpred P[set,set] means for A1 be Matrix of L,W,K st A1=$1 holds
Space_of_Solutions_of A9=Space_of_Solutions_of A1;
deffunc F(Matrix of L,W,K,Nat,Nat,Element of K) =$1;
A4: W>0 by A3,MATRIX13:74;
A5: for A1 be Matrix of L,W,K,B1 be Matrix of L,W,K st P[A1,B1] for a be
Element of K for i,j st j in dom A1 & (i = j implies a <> -1_K ) holds P[RLine(
A1,i,Line(A1,i) + a*Line(A1,j)),F(B1,i,j,a)]
proof
let A1 be Matrix of L,W,K,B1 be Matrix of L,W,K such that
A6: P[A1,B1];
let a be Element of K;
A7: dom A1 = Seg len A1 by FINSEQ_1:def 3
.= Seg L by MATRIX_0:def 2;
let i,j;
assume j in dom A1 &( i = j implies a <> -1_K);
then Space_of_Solutions_of RLine(A1,i,Line(A1,i) + a*Line(A1,j)) =
Space_of_Solutions_of A1 by A4,A7,Th65
.= Space_of_Solutions_of A9 by A6;
hence thesis;
end;
A8: P[A9,A9];
consider A1 be Matrix of L,W,K,B1 be Matrix of L,W,K,N such that
A9: N c= Seg W and
A10: the_rank_of A9 = the_rank_of A1 & the_rank_of A9 = card N and
A11: P[A1,B1] & Segm(A1,Seg card N,N) = 1.(K,card N) and
A12: for i st i in dom A1 & i > card N holds Line(A1,i) = W|->0.K and
for i,j st i in Seg card N & j in Seg width A1 & j < Sgm N.i holds A1*
(i,j) = 0.K from GAUSS2(A8,A5);
A13: 0 card N;
i in dom A1 by A24,XBOOLE_0:def 5;
hence Line(A1,i) = width A1 |-> 0.K by A12,A22,A14,A25,FINSEQ_1:1;
end;
card N <= len A1 by A10,MATRIX13:74;
then
A26: Seg card N c= Seg len A1 by FINSEQ_1:5;
width A1>0 by A4,A13,MATRIX_0:23;
then
A27: Space_of_Solutions_of SN = Space_of_Solutions_of A1 by A3,A10,A26,A22,A23
,Th66
.= Space_of_Solutions_of A9 by A11;
per cases;
suppose
A28: W-'card N=0;
then SN=1.(K,card N) by A9,A11,A14,A21,A18,CARD_2:102;
hence thesis by A10,A27,A18,A28,Lm8;
end;
suppose
A29: W-'card N>0;
then W-'card N+0>0;
then W-'card N>=1 by NAT_1:19;
then
A30: Det 1.(K,W-'card N)=1_K by MATRIX_7:16;
A31: card Seg (W-'card N)=W-'card N & 0.K<>1_K by FINSEQ_1:57;
consider MVectors be Matrix of W-'card N,W,K such that
A32: Segm(MVectors,Seg (W-'card N),Seg W \ N) = 1.(K,W-'card N) and
Segm(MVectors,Seg (W-'card N),N)=-Segm(SN,Seg card N,Seg W \N)@ and
A33: Lin(lines MVectors) = Space_of_Solutions_of A9 by A3,A9,A10,A27,A21,A15
,A20,A29,Th67;
len MVectors=W-'card N by A29,MATRIX_0:23;
then
A34: W-'card N>=the_rank_of MVectors by MATRIX13:74;
A35: Seg W \ N c= Seg W by XBOOLE_1:36;
Indices MVectors=[:Seg (W-'card N),Seg W:] by A29,MATRIX_0:23;
then
A36: [:Seg (W-'card N),Seg W \ N:] c= Indices MVectors by A35,ZFMISC_1:95;
reconsider B=lines MVectors as Subset of W-VectSp_over K;
A37: dom MVectors=Seg len MVectors & len MVectors=W-'card N by FINSEQ_1:def 3
,MATRIX_0:def 2;
EqSegm(MVectors,Seg (W-'card N),Seg W \ N)=1.(K,W-'card N) by A16,A18,A32
,FINSEQ_1:57,MATRIX13:def 3;
then
A38: W-'card N<=the_rank_of MVectors by A16,A18,A36,A30,A31,MATRIX13:def 4;
then MVectors is without_repeated_line by A34,MATRIX13:105,XXREAL_0:1;
then Seg (W-'card N),B are_equipotent by A37,WELLORD2:def 4;
then
A39: card B = card Seg (W-'card N) by CARD_1:5
.= W-'card N by FINSEQ_1:57;
the_rank_of MVectors = W-'card N by A38,A34,XXREAL_0:1;
then lines MVectors is linearly-independent by MATRIX13:121;
hence thesis by A10,A18,A33,A39,VECTSP_9:26;
end;
end;
end;
theorem Th69:
for M be Matrix of n,m,K, i,j,a st M is without_repeated_line &
j in dom M & (i = j implies a<>-1_K) holds Lin lines M = Lin lines RLine(M,i,
Line(M,i)+a*Line(M,j))
proof
let M be Matrix of n,m,K, i,j,a such that
A1: M is without_repeated_line and
A2: j in dom M and
A3: i = j implies a<>-1_K;
A4: len M=n by MATRIX_0:def 2;
set L=Line(M,i)+a*Line(M,j);
A5: dom M=Seg len M by FINSEQ_1:def 3;
set R=RLine(M,i,L);
per cases;
suppose
not i in dom M;
hence thesis by A5,MATRIX13:40;
end;
suppose
A6: i in dom M;
then n<>0 by A5,A4;
then
A7: width M=m by MATRIX_0:23;
then reconsider
Li=Line(M,i),Lj=Line(M,j) as Vector of m -VectSp_over K by MATRIX13:102;
a*Lj=a*Line(M,j) by A7,MATRIX13:102;
then
A8: L=Li+a*Lj by A7,MATRIX13:102;
A9: Li = Lj implies (a <> -1_K or Li = 0.(m -VectSp_over K))
proof
assume
A10: Li=Lj;
Li=M.i & Lj=M.j by A2,A5,A4,A6,MATRIX_0:52;
hence thesis by A1,A2,A3,A6,A10,FUNCT_1:def 4;
end;
reconsider L9=L as Element of (the carrier of K)* by FINSEQ_1:def 11;
reconsider LL=L9 as set;
set iL={i}-->L9;
len L=width M by CARD_1:def 7;
then
A11: R = M+*(i,LL) by MATRIX11:29
.= M+*(i.-->LL) by A6,FUNCT_7:def 3
.= M+*iL by FUNCOP_1:def 9;
M.:(dom M\dom iL) = M.:(dom M)\M.:(dom iL) by A1,FUNCT_1:64
.= rng M \ M.:(dom iL) by RELAT_1:113
.= rng M \ Im(M,i) by FUNCOP_1:13
.= rng M \ {M.i} by A6,FUNCT_1:59
.= rng M \ {Line(M,i)} by A5,A4,A6,MATRIX_0:52;
then
A12: lines R = lines M \ {Line(M,i)} \/rng iL by A11,FRECHET:12
.= lines M \ {Line(M,i)} \/{L} by FUNCOP_1:8;
A13: Lj in lines M by A2,A5,A4,MATRIX13:103;
Li in lines M by A5,A4,A6,MATRIX13:103;
hence thesis by A8,A12,A13,A9,Th14;
end;
end;
theorem Th70:
for W be Subspace of m-VectSp_over K ex A be Matrix of dim W,m,K
, N be without_zero finite Subset of NAT st N c= Seg m & dim W = card N & Segm(
A,Seg dim W,N)=1.(K,dim W) & the_rank_of A = dim W & lines A is Basis of W
proof
let W be Subspace of m-VectSp_over K;
consider I be finite Subset of W such that
A1: I is Basis of W by MATRLIN:def 1;
I is linearly-independent by A1,VECTSP_7:def 3;
then reconsider U=I as linearly-independent Subset of m-VectSp_over K by
VECTSP_9:11;
defpred P[set,set] means for A be Matrix of card I,m,K,B be Matrix of card I
,m,K st $1=A holds A is without_repeated_line & lines A is linearly-independent
& Lin(lines A)=(Omega).W;
deffunc F(Matrix of card I,m,K,Nat,Nat,Element of K) = $1;
consider M be Matrix of card I,m,K such that
A2: M is without_repeated_line & lines M = U by MATRIX13:104;
A3: for A9 be Matrix of card I,m,K, B9 be Matrix of card I,m,K st P[A9,B9]
for a be Element of K for i,j st j in dom A9 & (i=j implies a<>-1_K) holds P[
RLine(A9,i,Line(A9,i) + a*Line(A9,j)),F(B9,i,j,a)]
proof
let A9 be Matrix of card I,m,K, B9 be Matrix of card I,m,K such that
A4: P[A9,B9];
A5: dom A9=Seg len A9 by FINSEQ_1:def 3;
let a be Element of K;
let i,j such that
A6: j in dom A9 &( i = j implies a <> -1_K);
set R=RLine(A9,i,Line(A9,i) + a*Line(A9,j));
A7: A9 is without_repeated_line by A4;
then
A8: Lin lines A9= Lin lines R by A6,Th69;
lines A9 is linearly-independent by A4;
then card I = the_rank_of A9 by A7,MATRIX13:121
.= the_rank_of R by A6,A5,MATRIX13:92;
hence thesis by A4,A8,MATRIX13:121;
end;
Lin(I) = the ModuleStr of W by A1,VECTSP_7:def 3;
then
A9: P[M,M] by A2,VECTSP_9:17;
consider A9 be Matrix of card I,m,K, B9 be Matrix of card I,m,K,N such that
A10: N c= Seg m and
A11: the_rank_of M = the_rank_of A9 & the_rank_of M = card N & P[A9,B9] and
A12: Segm(A9,Seg card N,N) = 1.(K,card N) and
for i st i in dom A9 & i > card N holds Line(A9,i) = m|->0.K and
for i,j st i in Seg card N & j in Seg width A9 & j < Sgm N.i holds A9*(i
,j) = 0.K from GAUSS2(A9,A3);
dim W = card I by A1,VECTSP_9:def 1;
then reconsider A9 as Matrix of dim W,m,K;
lines A9 c= the carrier of Lin(lines A9)
by VECTSP_7:8,STRUCT_0:def 5;
then reconsider lA=lines A9 as linearly-independent Subset of W by A11,
VECTSP_9:12;
take A9,N;
A13: Lin(lA) = Lin(lines A9) by VECTSP_9:17;
A14: the_rank_of M = card I by A2,MATRIX13:121;
A15: card I=dim W by A1,VECTSP_9:def 1;
Lin(lines A9) = the ModuleStr of W by A11;
hence thesis by A15,A10,A11,A12,A14,A13,VECTSP_7:def 3;
end;
theorem
for W be strict Subspace of m-VectSp_over K st dim W < m ex A be
Matrix of m-'dim W,m,K, N be without_zero finite Subset of NAT st card N = m-'
dim W & N c= Seg m & Segm(A,Seg(m-'dim W),N) = 1.(K,m-'dim W) & W =
Space_of_Solutions_of A
proof
let W be strict Subspace of m-VectSp_over K such that
A1: dim W < m;
per cases;
suppose
A2: dim W=0;
then m-'dim W = m by NAT_D:40;
then reconsider ONE=1.(K,m) as Matrix of m-'dim W,m,K;
take ONE,Seg m;
A3: len 1.(K,m)=m by MATRIX_0:24;
A4: dim Space_of_Solutions_of ONE=0 by Lm8;
A5: m-'dim W=m by A2,NAT_D:40;
A6: width 1.(K,m)=m by MATRIX_0:24;
Space_of_Solutions_of ONE = (Omega).(Space_of_Solutions_of ONE)
.= (0).(Space_of_Solutions_of ONE) by A4,VECTSP_9:29
.= (0).W by A6,VECTSP_4:37
.= (Omega).W by A2,VECTSP_9:29
.= W;
hence thesis by A5,A3,A6,FINSEQ_1:57,MATRIX13:46;
end;
suppose
A7: dim W > 0;
set ZERO=0.(K,m-'dim W,m);
A8: m-dim W>dim W-dim W by A1,XREAL_1:9;
A9: m-'dim W=m-dim W by A1,XREAL_1:233;
then
A10: len ZERO=m-'dim W & width ZERO=m by A8,MATRIX_0:23;
A11: card Seg m = m by FINSEQ_1:57;
consider A be Matrix of dim W,m,K, N such that
A12: N c= Seg m and
A13: dim W = card N and
A14: Segm(A,Seg dim W,N)=1.(K,dim W) and
the_rank_of A = dim W and
A15: lines A is Basis of W by Th70;
set SA = Segm(A,Seg dim W,(Seg m)\N);
A16: card ((Seg m)\N) = card Seg m-card N by A12,CARD_2:44;
then
A17: width SA =m-card N by A7,A11,MATRIX_0:23;
A18: card Seg dim W=dim W by FINSEQ_1:57;
then len SA=dim W by A7,MATRIX_0:23;
then width (SA@) =dim W by A13,A8,A17,MATRIX_0:54;
then
A19: width (-SA@) =dim W by MATRIX_3:def 2;
A20: card Seg (m-'dim W)=m-'dim W by FINSEQ_1:57;
m-'dim W = m-dim W by A1,XREAL_1:233;
then reconsider
CC=1.(K,m-'dim W) as Matrix of card Seg(m-'dim W),card((Seg m)\
N),K by A13,A16,A11,A20;
A21: Seg m\((Seg m)\N) = Seg m /\N & m-'(m-'dim W)=m-(m-'dim W) by NAT_D:35
,XBOOLE_1:48,XREAL_1:233;
A22: Indices ZERO=[:Seg (m-'dim W ), Seg m:] by A9,A8,MATRIX_0:23;
then
A23: [:Seg (m-'dim W),N:] c= Indices ZERO by A12,ZFMISC_1:95;
len (SA@)=m-dim W by A13,A8,A17,MATRIX_0:54;
then len (-SA@)=m-'dim W by A9,MATRIX_3:def 2;
then reconsider
BB=-SA@ as Matrix of card Seg (m-'dim W),card N,K by A13,A20,A19,
MATRIX_0:51;
A24: N misses (Seg m)\N by XBOOLE_1:79;
A25: (Seg m)\N c= Seg m by XBOOLE_1:36;
then
A26: [:Seg (m-'dim W),(Seg m)\N:] c= Indices ZERO by A22,ZFMISC_1:95;
[:Seg (m-'dim W),N:]/\[:Seg (m-'dim W),(Seg m)\N:]= [:Seg (m-'dim W),
N/\((Seg m)\N):] by ZFMISC_1:99
.= [:Seg (m-'dim W),{}:] by A24
.= {} by ZFMISC_1:90;
then
for i,j,bi,bj,ci,cj be Nat st [i,j] in [:Seg (m-'dim W),N:]/\[:Seg (m
-'dim W),(Seg m)\N:] & bi=Sgm Seg(m-'dim W)".i & bj=Sgm N".j & ci=Sgm Seg(m-'
dim W)".i& cj=Sgm ((Seg m)\N)".j holds BB*(bi,bj) = CC*(ci,cj);
then consider M be Matrix of m-'dim W,m,K such that
A27: Segm(M,Seg (m-'dim W),N)=BB and
A28: Segm(M,Seg (m-'dim W),(Seg m)\N)=CC and
for i,j st [i,j] in Indices M\([:Seg (m-'dim W),N:]\/[:Seg (m-'dim W),
(Seg m)\N:]) holds M*(i,j) = ZERO*(i,j) by A10,A23,A26,Th9;
Seg m /\ N =N by A12,XBOOLE_1:28;
then consider MV be Matrix of dim W,m,K such that
A29: Segm(MV,Seg dim W,N) = 1.(K,dim W) and
A30: Segm(MV,Seg dim W,(Seg m)\N)= -Segm(M,Seg (m-'dim W),N)@ and
A31: Lin(lines MV) = Space_of_Solutions_of M by A7,A13,A9,A8,A16,A11,A28,A21
,Th67,XBOOLE_1:36;
A32: now
A33: Indices A=[:Seg dim W, Seg m:] by A7,MATRIX_0:23;
let i,j such that
A34: [i,j] in Indices A;
A35: i in Seg dim W by A34,A33,ZFMISC_1:87;
A36: Indices A=Indices MV by MATRIX_0:26;
A37: rng Sgm Seg dim W=Seg dim W by FINSEQ_1:def 13;
dom Sgm Seg dim W=Seg dim W by A18,FINSEQ_3:40;
then consider x being object such that
A38: x in Seg dim W and
A39: (Sgm Seg dim W).x=i by A35,A37,FUNCT_1:def 3;
reconsider x as Element of NAT by A38;
A40: j in Seg m by A34,A33,ZFMISC_1:87;
now
per cases;
suppose
A41: j in N;
then
A42: [i,j] in [:Seg dim W,N:] by A35,ZFMISC_1:87;
A43: rng Sgm N=N by A12,FINSEQ_1:def 13;
dom Sgm N=Seg dim W by A12,A13,FINSEQ_3:40;
then consider y being object such that
A44: y in Seg dim W and
A45: (Sgm N).y=j by A41,A43,FUNCT_1:def 3;
reconsider y as Element of NAT by A44;
A46: [:Seg dim W,N:] c= Indices A by A12,A33,ZFMISC_1:95;
then
A47: [x,y] in Indices Segm(MV,Seg dim W,N) by A36,A37,A39,A43,A45,A42,
MATRIX13:17;
[x,y] in Indices Segm(A,Seg dim W,N) by A37,A39,A43,A45,A46,A42,
MATRIX13:17;
hence A*(i,j) = Segm(MV,Seg dim W,N)*(x,y) by A14,A29,A39,A45,
MATRIX13:def 1
.= MV*(i,j) by A39,A45,A47,MATRIX13:def 1;
end;
suppose
not j in N;
then
A48: j in Seg m\N by A40,XBOOLE_0:def 5;
then
A49: [i,j] in [:Seg dim W,Seg m\N:] by A35,ZFMISC_1:87;
A50: rng Sgm (Seg m\N)=(Seg m\N) by A25,FINSEQ_1:def 13;
dom Sgm(Seg m\N)=Seg(m-'dim W) by A13,A9,A16,A11,FINSEQ_3:40
,XBOOLE_1:36;
then consider y being object such that
A51: y in Seg (m-'dim W) and
A52: (Sgm (Seg m\N)).y=j by A48,A50,FUNCT_1:def 3;
reconsider y as Element of NAT by A51;
A53: [:Seg dim W,Seg m\N:] c= Indices A by A25,A33,ZFMISC_1:95;
then
A54: [x,y] in Indices Segm(A,Seg dim W,Seg m\N) by A37,A39,A50,A52,A49,
MATRIX13:17;
A55: [x,y] in Indices Segm(MV,Seg dim W,Seg m\N) by A36,A37,A39,A50,A52
,A53,A49,MATRIX13:17;
then
A56: [x,y] in Indices (Segm(M,Seg (m-'dim W),N)@) by A30,Lm1;
then
A57: [y,x] in Indices Segm(M,Seg (m-'dim W),N) by MATRIX_0:def 6;
then
A58: [y,x] in Indices SA@ by A27,Lm1;
thus MV*(i,j) = (-Segm(M,Seg (m-'dim W),N)@)*(x,y) by A30,A39,A52,A55
,MATRIX13:def 1
.= -(Segm(M,Seg (m-'dim W),N)@)*(x,y) by A56,MATRIX_3:def 2
.= -(-SA@)*(y,x) by A27,A57,MATRIX_0:def 6
.= --(SA@)*(y,x) by A58,MATRIX_3:def 2
.= (SA@)*(y,x) by RLVECT_1:17
.= SA*(x,y) by A54,MATRIX_0:def 6
.=A*(i,j) by A39,A52,A54,MATRIX13:def 1;
end;
end;
hence A*(i,j)=MV*(i,j);
end;
then reconsider lA=lines MV as Subset of W by A15,MATRIX_0:27;
take M,Seg m\ N;
MV = A by A32,MATRIX_0:27;
then Lin(lA) = the ModuleStr of W by A15,VECTSP_7:def 3;
hence thesis by A1,A13,A16,A11,A28,A31,VECTSP_9:17,XBOOLE_1:36,XREAL_1:233;
end;
end;
theorem Th72:
for A,B be Matrix of K st width A = len B & ( width A = 0
implies len A = 0 ) & ( width B = 0 implies len B = 0 ) holds
Space_of_Solutions_of B is Subspace of Space_of_Solutions_of (A*B)
proof
let A,B be Matrix of K such that
A1: width A = len B and
A2: width A = 0 implies len A = 0 and
A3: width B = 0 implies len B = 0;
set AB=A*B;
A4: len AB=len A by A1,MATRIX_3:def 4;
A5: width AB=width B by A1,MATRIX_3:def 4;
then reconsider AB as Matrix of len A,width B,K by A4,MATRIX_0:51;
the carrier of Space_of_Solutions_of B c= the carrier of
Space_of_Solutions_of AB
proof
let x be object;
assume x in the carrier of Space_of_Solutions_of B;
then x in Solutions_of(B,len B|->0.K) by A3,Def5;
then consider f such that
A6: f=x and
A7: ColVec2Mx f in Solutions_of(B,ColVec2Mx (len B|->0.K));
consider X such that
A8: X=ColVec2Mx f and
A9: len X = width B and
A10: width X = width ColVec2Mx (len B|->0.K) and
A11: B * X = ColVec2Mx (len B|->0.K) by A7;
A12: ColVec2Mx (len AB|->0.K)=0.(K,len A,1) by A4,Th32;
A13: ColVec2Mx (len B|->0.K)=0.(K,len B,1) by Th32;
now
per cases;
suppose
len A=0;
then
Solutions_of(AB,ColVec2Mx (len AB|->0.K)) ={{}} & X ={} by A4,A5,A9,A12
,Th51,MATRIX_0:def 3;
hence X in Solutions_of(AB,ColVec2Mx (len AB|->0.K)) by TARSKI:def 1;
end;
suppose
A14: len A<>0;
then
A15: width ColVec2Mx (len AB|->0.K)=1 by A4,Th26
.= width ColVec2Mx (len B|->0.K) by A1,A2,A14,Th26;
ColVec2Mx (len AB|->0.K) = A*(B*X) by A1,A2,A11,A13,A12,A14,MATRIXR2:18
.= AB*X by A1,A9,MATRIX_3:33;
hence X in Solutions_of(AB,ColVec2Mx (len AB|->0.K)) by A5,A9,A10,A15;
end;
end;
then f in Solutions_of(AB,len AB|->0.K) by A8;
hence thesis by A1,A2,A3,A4,A5,A6,Def5;
end;
hence thesis by A5,VECTSP_4:27;
end;
theorem
for A,B be Matrix of K st width A = len B holds the_rank_of (A*B) <=
the_rank_of A & the_rank_of (A*B) <= the_rank_of B
proof
let A,B be Matrix of K such that
A1: width A = len B;
set AB=A*B;
A2: width AB=width B by A1,MATRIX_3:def 4;
per cases;
suppose
the_rank_of AB=0;
hence thesis;
end;
suppose
A3: the_rank_of AB>0;
set AT=A@;
A4: width AB>0 by A3,MATRIX13:74;
then
A5: width A>0 by A1,A2,MATRIX_0:def 3;
then
A6: len AT=width A by MATRIX_0:54;
set BT=B@;
set BA=BT*AT;
width AT=len A by A5,MATRIX_0:54;
then
A7: width AT = 0 implies len AT = 0 by A5,MATRIX_0:def 3;
then
A8: dim Space_of_Solutions_of AT = width AT - the_rank_of AT by Th68;
A9: width BT=len B by A2,A4,MATRIX_0:54;
then width BT = 0 implies len BT = 0 by A2,A4,MATRIX_0:def 3;
then
A10: Space_of_Solutions_of AT is Subspace of Space_of_Solutions_of BA by A1,A6
,A9,A7,Th72;
A11: width BA=width AT by A1,A6,A9,MATRIX_3:def 4;
then dim Space_of_Solutions_of BA = width BA - the_rank_of BA by A5,A7,Th68
,MATRIX_0:54;
then width AT - the_rank_of AT<=width AT - the_rank_of BA by A11,A10,A8,
VECTSP_9:25;
then the_rank_of AT >= the_rank_of BA by XREAL_1:10;
then
A12: the_rank_of A >= the_rank_of BA by MATRIX13:84;
width A=0 implies len A=0 by A1,A2,A4,MATRIX_0:def 3;
then
A13: Space_of_Solutions_of B is Subspace of Space_of_Solutions_of AB by A1,A2
,A4,Th72;
dim Space_of_Solutions_of B = width B-the_rank_of B & dim
Space_of_Solutions_of AB=width AB-the_rank_of AB by A2,A4,Th68;
then
A14: width B-the_rank_of B <= width B-the_rank_of AB by A2,A13,VECTSP_9:25;
BA=AB@ by A1,A2,A4,MATRIX_3:22;
hence thesis by A14,A12,MATRIX13:84,XREAL_1:10;
end;
end;
theorem Th74:
for A be Matrix of n,n,K, B be Matrix of K st Det A <> 0.K &
width A = len B & ( width B = 0 implies len B = 0 ) holds Space_of_Solutions_of
B = Space_of_Solutions_of (A*B)
proof
let A be Matrix of n,n,K,B be Matrix of K such that
A1: Det A <> 0.K and
A2: width A=len B and
A3: width B = 0 implies len B = 0;
set AB=A*B;
A4: len AB=len A by A2,MATRIX_3:def 4;
A5: width AB=width B by A2,MATRIX_3:def 4;
A6: len A=n by MATRIX_0:24;
reconsider AB as Matrix of n,width B,K by A4,A5,MATRIX_0:24,51;
A7: width A=n by MATRIX_0:24;
A8: the carrier of Space_of_Solutions_of AB c= the carrier of
Space_of_Solutions_of B
proof
A is invertible by A1,LAPLACE:34;
then A is_reverse_of A~ by MATRIX_6:def 4;
then
A9: 1.(K,n)=A~*A by MATRIX_6:def 2;
A10: len (A~)=n by MATRIX_0:24;
let x be object;
assume x in the carrier of Space_of_Solutions_of AB;
then x in Solutions_of(AB,len AB|->0.K) by A2,A3,A6,A7,A4,A5,Def5;
then consider f such that
A11: f=x and
A12: ColVec2Mx f in Solutions_of(AB,ColVec2Mx (len AB|->0.K));
consider X such that
A13: X=ColVec2Mx f and
A14: len X = width AB and
A15: width X = width ColVec2Mx (len AB|->0.K) and
A16: AB * X = ColVec2Mx (len AB|->0.K) by A12;
A17: width (A~)=n by MATRIX_0:24;
set BX=B*X;
A18: len BX=len B by A5,A14,MATRIX_3:def 4;
then
A19: BX = 1.(K,n)* BX by A2,A7,MATRIXR2:68
.= A~ * (A*BX) by A2,A6,A9,A17,A18,MATRIX_3:33
.= A~ * ColVec2Mx(len AB|->0.K) by A2,A5,A14,A16,MATRIX_3:33
.= A~ * 0.(K,len AB,1) by Th32;
now
per cases;
suppose
A20: n = 0;
then 0.(K,len AB,1)={} by A6,A4;
hence BX=0.(K,len AB,1) by A2,A18,A20,MATRIX_0:24;
end;
suppose
n>0;
hence BX=0.(K,len AB,1) by A6,A4,A10,A17,A19,MATRIXR2:18;
end;
end;
then BX =ColVec2Mx(len AB|->0.K) by Th32;
then
ColVec2Mx f in Solutions_of(B,ColVec2Mx(len B|->0.K)) by A2,A6,A7,A4,A5,A13
,A14,A15;
then f in Solutions_of(B,len B|->0.K);
hence thesis by A3,A11,Def5;
end;
width A=0 implies len A=0 by A6,MATRIX_0:24;
then Space_of_Solutions_of B is Subspace of Space_of_Solutions_of (A*B) by A2
,A3,Th72;
then the carrier of Space_of_Solutions_of B c= the carrier of
Space_of_Solutions_of (A*B) by VECTSP_4:def 2;
then the carrier of Space_of_Solutions_of B = the carrier of
Space_of_Solutions_of (A*B) by A8;
hence thesis by A5,VECTSP_4:29;
end;
theorem Th75:
for A be Matrix of n,n,K, B be Matrix of K st width A = len B &
Det A <> 0.K holds the_rank_of (A*B) = the_rank_of B
proof
let A be Matrix of n,n,K, B be Matrix of K such that
A1: width A = len B and
A2: Det A <>0.K;
set AB=A*B;
A3: len AB=len A by A1,MATRIX_3:def 4;
A4: len A=n & width A=n by MATRIX_0:24;
A5: width AB=width B by A1,MATRIX_3:def 4;
per cases;
suppose
width AB=0;
hence thesis by A1,A3,A5,A4,Lm3;
end;
suppose
A6: width AB>0;
then Space_of_Solutions_of B = Space_of_Solutions_of AB & dim
Space_of_Solutions_of B = width B-the_rank_of B by A1,A2,A5,Th68,Th74;
then width B-the_rank_of B=width B-the_rank_of AB by A5,A6,Th68;
hence thesis;
end;
end;
theorem
for A be Matrix of n,n,K, B be Matrix of K st len A = width B & Det A
<> 0.K holds the_rank_of (B*A) = the_rank_of B
proof
let A be Matrix of n,n,K,B be Matrix of K such that
A1: width B = len A and
A2: Det A <> 0.K;
set BA=B*A;
A3: len BA=len B by A1,MATRIX_3:def 4;
A4: width BA=width A by A1,MATRIX_3:def 4;
A5: len A=n & width A=n by MATRIX_0:24;
per cases;
suppose
width BA=0;
hence thesis by A1,A3,A4,A5,Lm3;
end;
suppose
A6: width BA>0;
then
A7: width (A@)=len A & len (B@)=width B by A1,A4,A5,MATRIX_0:54;
A8: Det (A@)<>0.K by A2,MATRIXR2:43;
thus the_rank_of (B*A) = the_rank_of ((B*A)@) by MATRIX13:84
.= the_rank_of ((A@)*(B@)) by A1,A4,A6,MATRIX_3:22
.= the_rank_of (B@) by A1,A8,A7,Th75
.= the_rank_of B by MATRIX13:84;
end;
end;