let n be Nat; for K being Field
for M1, M2 being Matrix of n,K st M2 is invertible & M1 is_congruent_Matrix_of M2 & n > 0 holds
M1 is invertible
let K be Field; for M1, M2 being Matrix of n,K st M2 is invertible & M1 is_congruent_Matrix_of M2 & n > 0 holds
M1 is invertible
let M1, M2 be Matrix of n,K; ( M2 is invertible & M1 is_congruent_Matrix_of M2 & n > 0 implies M1 is invertible )
assume that
A1:
M2 is invertible
and
A2:
M1 is_congruent_Matrix_of M2
and
A3:
n > 0
; M1 is invertible
consider M4 being Matrix of n,K such that
A4:
M4 is invertible
and
A5:
M1 = ((M4 @) * M2) * M4
by A2;
set M6 = (M4 ~) @ ;
set M5 = M4 @ ;
A6:
( width M4 = n & width (M4 ~) = n )
by MATRIX_0:24;
len M4 = n
by MATRIX_0:24;
then A7:
((M4 ~) * M4) @ = (M4 @) * ((M4 ~) @)
by A3, A6, MATRIX_3:22;
A8:
M4 ~ is_reverse_of M4
by A4, MATRIX_6:def 4;
then
(M4 ~) * M4 = 1. (K,n)
by MATRIX_6:def 2;
then A9:
(M4 @) * ((M4 ~) @) = 1. (K,n)
by A7, MATRIX_6:10;
len (M4 ~) = n
by MATRIX_0:24;
then
(M4 * (M4 ~)) @ = ((M4 ~) @) * (M4 @)
by A3, A6, MATRIX_3:22;
then
(M4 @) * ((M4 ~) @) = ((M4 ~) @) * (M4 @)
by A8, A7, MATRIX_6:def 2;
then
M4 @ is_reverse_of (M4 ~) @
by A9, MATRIX_6:def 2;
then
M4 @ is invertible
by MATRIX_6:def 3;
then
(M4 @) * M2 is invertible
by A1, MATRIX_6:36;
hence
M1 is invertible
by A4, A5, MATRIX_6:36; verum