let m be Nat; for K being Field
for A, B being Matrix of K st B = 0. (K,(len A),m) holds
the_rank_of A = the_rank_of (A ^^ B)
let K be Field; for A, B being Matrix of K st B = 0. (K,(len A),m) holds
the_rank_of A = the_rank_of (A ^^ B)
let A, B be Matrix of K; ( B = 0. (K,(len A),m) implies the_rank_of A = the_rank_of (A ^^ B) )
assume A1:
B = 0. (K,(len A),m)
; the_rank_of A = the_rank_of (A ^^ B)
A2:
len B = len A
by A1, MATRIX_0:def 2;
set L = len A;
reconsider B9 = B as Matrix of len A, width B,K by A2, MATRIX_0:51;
reconsider A9 = A as Matrix of len A, width A,K by MATRIX_0:51;
set AB = A9 ^^ B9;
per cases
( width B = 0 or width B > 0 )
;
suppose
width B > 0
;
the_rank_of A = the_rank_of (A ^^ B)then A3:
len A > 0
by A2, MATRIX_0:def 3;
then A4:
width (A9 ^^ B9) = (width A) + (width B)
by MATRIX_0:23;
A5:
len (A9 ^^ B9) = len A
by A3, MATRIX_0:23;
A6:
now for i being Nat st i in (Seg (width (A9 ^^ B9))) \ (Seg (width A)) holds
Col ((A9 ^^ B9),i) = (len (A9 ^^ B9)) |-> (0. K)set L0 =
(len (A9 ^^ B9)) |-> (0. K);
let i be
Nat;
( i in (Seg (width (A9 ^^ B9))) \ (Seg (width A)) implies Col ((A9 ^^ B9),i) = (len (A9 ^^ B9)) |-> (0. K) )assume A7:
i in (Seg (width (A9 ^^ B9))) \ (Seg (width A))
;
Col ((A9 ^^ B9),i) = (len (A9 ^^ B9)) |-> (0. K)A8:
i in Seg (width (A9 ^^ B9))
by A7, XBOOLE_0:def 5;
not
i in Seg (width A)
by A7, XBOOLE_0:def 5;
then A9:
(
i < 1 or
i > width A )
;
then reconsider n =
i - (width A) as
Element of
NAT by A8, FINSEQ_1:1, NAT_1:21;
A10:
i = n + (width A)
;
n <> 0
by A8, A9, FINSEQ_1:1;
then A11:
n in Seg (width B)
by A4, A8, A10, FINSEQ_1:61;
A12:
now for i being Nat st 1 <= i & i <= len A holds
(Col (B9,n)) . i = ((len (A9 ^^ B9)) |-> (0. K)) . ilet i be
Nat;
( 1 <= i & i <= len A implies (Col (B9,n)) . i = ((len (A9 ^^ B9)) |-> (0. K)) . i )assume A13:
( 1
<= i &
i <= len A )
;
(Col (B9,n)) . i = ((len (A9 ^^ B9)) |-> (0. K)) . iA14:
i in Seg (len A)
by A13;
then A15:
((len (A9 ^^ B9)) |-> (0. K)) . i = 0. K
by A5, FINSEQ_2:57;
Seg (len A) = dom B
by A2, FINSEQ_1:def 3;
then
(
(Col (B9,n)) . i = B9 * (
i,
n) &
[i,n] in Indices B )
by A11, A14, MATRIX_0:def 8, ZFMISC_1:87;
hence
(Col (B9,n)) . i = ((len (A9 ^^ B9)) |-> (0. K)) . i
by A1, A15, MATRIX_3:1;
verum end; A16:
(
len (Col (B,n)) = len A &
len ((len (A9 ^^ B9)) |-> (0. K)) = len A )
by A2, A5, CARD_1:def 7;
Col (
(A9 ^^ B9),
i)
= Col (
B9,
n)
by A10, A11, Th17;
hence
Col (
(A9 ^^ B9),
i)
= (len (A9 ^^ B9)) |-> (0. K)
by A16, A12;
verum end;
width A <= width (A9 ^^ B9)
by A4, NAT_1:11;
then
Seg (width A) c= Seg (width (A9 ^^ B9))
by FINSEQ_1:5;
hence the_rank_of (A ^^ B) =
the_rank_of (Segm ((A9 ^^ B9),(Seg (len (A9 ^^ B9))),(Seg (width A))))
by A6, Th12
.=
the_rank_of A
by A5, Th19
;
verum end; end;