let n, m be Nat; for K being Field st n > 0 & not Solutions_of (0. K,n,0 ),(0. K,n,m) is empty holds
m = 0
let K be Field; ( n > 0 & not Solutions_of (0. K,n,0 ),(0. K,n,m) is empty implies m = 0 )
assume that
A1:
n > 0
and
A2:
not Solutions_of (0. K,n,0 ),(0. K,n,m) is empty
; m = 0
consider x being set such that
A3:
x in Solutions_of (0. K,n,0 ),(0. K,n,m)
by A2, XBOOLE_0:def 1;
A4:
width (0. K,n,0 ) = 0
by A1, MATRIX_1:24;
ex X being Matrix of K st
( X = x & len X = width (0. K,n,0 ) & width X = width (0. K,n,m) & (0. K,n,0 ) * X = 0. K,n,m )
by A3;
hence 0 =
width (0. K,n,m)
by A4, MATRIX_1:def 4
.=
m
by A1, MATRIX_1:24
;
verum