reserve x,y for object,X,Y for set,
  D for non empty set,
  i,j,k,l,m,n,m9,n9 for Nat,
  i0,j0,n0,m0 for non zero Nat,
  K for Field,
  a,b for Element of K,
  p for FinSequence of K,
  M for Matrix of n,K;
reserve A for (Matrix of D),
  A9 for Matrix of n9,m9,D,
  M9 for Matrix of n9, m9,K,
  nt,nt1,nt2 for Element of n-tuples_on NAT,
  mt,mt1 for Element of m -tuples_on NAT,
  M for Matrix of K;
reserve P,P1,P2,Q,Q1,Q2 for without_zero finite Subset of NAT;

theorem
  a <> 0.K implies the_rank_of M = the_rank_of (a*M)
proof
  consider P,Q such that
A1: [:P,Q:] c= Indices M and
A2: card P = card Q and
A3: card P = the_rank_of M and
A4: Det EqSegm(M,P,Q)<>0.K by Def4;
  consider P1,Q1 such that
A5: [:P1,Q1:] c= Indices (a*M) and
A6: card P1 = card Q1 and
A7: card P1 = the_rank_of (a*M) and
A8: Det EqSegm(a*M,P1,Q1)<>0.K by Def4;
A9: Indices M = Indices (a*M) by MATRIXR1:18;
  then Det EqSegm(a*M,P1,Q1)=power(K).(a,card P1)*Det EqSegm(M,P1,Q1) by A5,A6
,Th72;
  then Det EqSegm(M,P1,Q1)<>0.K by A8;
  then
A10: the_rank_of M >= the_rank_of (a*M) by A9,A5,A6,A7,Def4;
  assume a<>0.K;
  then
A11: power(K).(a,card P)<>0.K by Lm6;
  Det EqSegm(a*M,P,Q)=power(K).(a,card P)*Det EqSegm(M,P,Q) by A1,A2,Th72;
  then Det EqSegm(a*M,P,Q)<>0.K by A4,A11,VECTSP_1:12;
  then the_rank_of (a*M) >= the_rank_of M by A9,A1,A2,A3,Def4;
  hence thesis by A10,XXREAL_0:1;
end;
