let n be Element of NAT ; for K being Field
for A being Matrix of n,K holds
( A = 0. (K,n) iff for i, j being Element of NAT st 1 <= i & i <= n & 1 <= j & j <= n holds
A * (i,j) = 0. K )
let K be Field; for A being Matrix of n,K holds
( A = 0. (K,n) iff for i, j being Element of NAT st 1 <= i & i <= n & 1 <= j & j <= n holds
A * (i,j) = 0. K )
let A be Matrix of n,K; ( A = 0. (K,n) iff for i, j being Element of NAT st 1 <= i & i <= n & 1 <= j & j <= n holds
A * (i,j) = 0. K )
A1:
width A = n
by MATRIX_0:24;
thus
( A = 0. (K,n) implies for i, j being Element of NAT st 1 <= i & i <= n & 1 <= j & j <= n holds
A * (i,j) = 0. K )
( ( for i, j being Element of NAT st 1 <= i & i <= n & 1 <= j & j <= n holds
A * (i,j) = 0. K ) implies A = 0. (K,n) )
assume A3:
for i, j being Element of NAT st 1 <= i & i <= n & 1 <= j & j <= n holds
A * (i,j) = 0. K
; A = 0. (K,n)
A4:
Indices A = [:(Seg n),(Seg n):]
by MATRIX_0:24;
A5:
for i, j being Nat st [i,j] in Indices A holds
A * (i,j) = (A + A) * (i,j)
proof
let i,
j be
Nat;
( [i,j] in Indices A implies A * (i,j) = (A + A) * (i,j) )
reconsider i0 =
i,
j0 =
j as
Element of
NAT by ORDINAL1:def 12;
assume A6:
[i,j] in Indices A
;
A * (i,j) = (A + A) * (i,j)
then
j in Seg n
by A1, ZFMISC_1:87;
then A7:
( 1
<= j &
j <= n )
by FINSEQ_1:1;
i in Seg n
by A4, A6, ZFMISC_1:87;
then
( 1
<= i &
i <= n )
by FINSEQ_1:1;
then
A * (
i0,
j0)
= 0. K
by A3, A7;
then (A + A) * (
i,
j) =
(0. K) + (A * (i,j))
by A6, MATRIX_3:def 3
.=
A * (
i,
j)
by RLVECT_1:4
;
hence
A * (
i,
j)
= (A + A) * (
i,
j)
;
verum
end;
len A = n
by MATRIX_0:24;
then
A = 0. (K,n,n)
by A1, A5, MATRIX_0:27, MATRIX_4:6;
hence
A = 0. (K,n)
by MATRIX_3:def 1; verum