let n be Nat; for R being Ring holds (1. (R,n)) @ = 1. (R,n)
let R be Ring; (1. (R,n)) @ = 1. (R,n)
per cases
( n > 0 or n = 0 )
by NAT_1:3;
suppose A1:
n > 0
;
(1. (R,n)) @ = 1. (R,n)A2:
len (1. (R,n)) = n
by MATRIX_0:24;
A3:
Indices (1. (R,n)) = [:(Seg n),(Seg n):]
by MATRIX_0:24;
A4:
for
i,
j being
Nat st
[i,j] in Indices (1. (R,n)) holds
(1. (R,n)) * (
i,
j)
= ((1. (R,n)) @) * (
i,
j)
proof
let i,
j be
Nat;
( [i,j] in Indices (1. (R,n)) implies (1. (R,n)) * (i,j) = ((1. (R,n)) @) * (i,j) )
assume A5:
[i,j] in Indices (1. (R,n))
;
(1. (R,n)) * (i,j) = ((1. (R,n)) @) * (i,j)
then
(
i in Seg n &
j in Seg n )
by A3, ZFMISC_1:87;
then A6:
[j,i] in Indices (1. (R,n))
by A3, ZFMISC_1:87;
per cases
( i = j or i <> j )
;
suppose
i <> j
;
(1. (R,n)) * (i,j) = ((1. (R,n)) @) * (i,j)then
(
(1. (R,n)) * (
i,
j)
= 0. R &
(1. (R,n)) * (
j,
i)
= 0. R )
by A5, A6, MATRIX_1:def 3;
hence
(1. (R,n)) * (
i,
j)
= ((1. (R,n)) @) * (
i,
j)
by A6, MATRIX_0:def 6;
verum end; end;
end; A7:
width (1. (R,n)) = n
by MATRIX_0:24;
then
(
len ((1. (R,n)) @) = width (1. (R,n)) &
width ((1. (R,n)) @) = len (1. (R,n)) )
by A1, MATRIX_0:54;
hence
(1. (R,n)) @ = 1. (
R,
n)
by A7, A2, A4, MATRIX_0:21;
verum end; end;