let K be Field; for A being Matrix of K st ( width A <> 0 or len A = 0 ) holds
not Solutions_of (A,((len A) |-> (0. K))) is empty
let A be Matrix of K; ( ( width A <> 0 or len A = 0 ) implies not Solutions_of (A,((len A) |-> (0. K))) is empty )
set L = (len A) |-> (0. K);
A1:
len ((len A) |-> (0. K)) = len A
by CARD_1:def 7;
reconsider A9 = A as Matrix of len A, width A,K by MATRIX_0:51;
assume A2:
( width A <> 0 or len A = 0 )
; not Solutions_of (A,((len A) |-> (0. K))) is empty
per cases
( len A = 0 or width A > 0 )
by A2;
suppose A5:
width A > 0
;
not Solutions_of (A,((len A) |-> (0. K))) is empty
ColVec2Mx ((len A) |-> (0. K)) = 0. (
K,
(len A),1)
by Th32;
then
(
len (ColVec2Mx ((len A) |-> (0. K))) = len ((len A) |-> (0. K)) &
the_rank_of A = the_rank_of (A ^^ (ColVec2Mx ((len A) |-> (0. K)))) )
by Th23, MATRIX_0:def 2;
then
not
Solutions_of (
A,
(ColVec2Mx ((len A) |-> (0. K)))) is
empty
by A1, A5, Th57;
then consider x being
object such that A6:
x in Solutions_of (
A,
(ColVec2Mx ((len A) |-> (0. K))))
;
consider f being
FinSequence of
K such that A7:
x = ColVec2Mx f
and
len f = width A
by A6, Th58;
f in Solutions_of (
A,
((len A) |-> (0. K)))
by A6, A7;
hence
not
Solutions_of (
A,
((len A) |-> (0. K))) is
empty
;
verum end; end;