theorem
  for V being finite-rank free Z_Module,
  b1, b2 being OrdBasis of V, f being bilinear-Form of V, V
  holds |. Det BilinearM(f, b2, b2) .| = |. Det BilinearM(f, b1, b1) .|
  proof
    let V be finite-rank free Z_Module,
    b1, b2 be OrdBasis of V, f be bilinear-Form of V, V;
    set n = len b1;
    A1: len b1 = rank V by ThRank1;
    A2: len b2 = rank V by ThRank1;
    reconsider B1 = BilinearM(f, b1, b1) as Matrix of n,INT.Ring;
    reconsider B2 = BilinearM(f, b2, b2) as Matrix of n,INT.Ring by A1,A2;
    per cases;
    suppose rank V = 0;
      hence |. Det BilinearM(f, b2, b2) .| = |. Det BilinearM(f, b1, b1) .|
      by A1,A2,MATRIX_0:45;
    end;
    suppose ZZ: rank V > 0;
      then B2: BilinearM(f, b2, b2) = ((AutMt(id(V), b2, b1))
      * BilinearM(f, b1, b1)) * (AutMt(id(V), b2, b1)@) by ThMBF3;
      reconsider IM1 = AutMt(id(V), b2, b1) as Matrix of n,INT.Ring
      by A1,LMThMBF3;
      reconsider IM2 = AutMt(id(V), b2, b1) as Matrix of n,INT.Ring
      by A1,LMThMBF3;
      reconsider M1 = IM1@ as Matrix of n,INT.Ring;
      reconsider M2 = IM2 as Matrix of n,INT.Ring;
      n >= 1+0 by A1,ZZ,NAT_1:13; then
      X1: Det IM1 = Det M1 by MATRIX_7:37;
      reconsider M2B1 = M2 * B1 as Matrix of n, INT.Ring;
      Det B2 = (Det(M2B1)) * (Det M1) by B2,MATRIX11:62,ZZ,A1
      .= ((Det M2) * (Det B1)) * (Det M1) by ZZ,A1,MATRIX11:62;
      hence |. Det BilinearM(f, b2, b2) .|
      = |. (Det M2) * (Det B1) .| * |. Det M1 .| by A1,A2,COMPLEX1:65
      .= |. (Det M2) * (Det B1) .| * 1 by A1,X1,ThSign1
      .= |. Det M2 .| * |. Det B1 .| by COMPLEX1:65
      .= |. Det B1 .| * 1 by A1,ThSign1
      .= |. Det BilinearM(f, b1, b1) .|;
    end;
  end;
