let K be Field; for M being Matrix of K
for RANK being Element of NAT holds
( the_rank_of M = RANK iff ( ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K ) & ( for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK ) ) )
let M be Matrix of K; for RANK being Element of NAT holds
( the_rank_of M = RANK iff ( ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K ) & ( for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK ) ) )
let RANK be Element of NAT ; ( the_rank_of M = RANK iff ( ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K ) & ( for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK ) ) )
thus
( the_rank_of M = RANK implies ( ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K ) & ( for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK ) ) )
( ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K ) & ( for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK ) implies the_rank_of M = RANK )proof
assume A1:
the_rank_of M = RANK
;
( ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K ) & ( for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK ) )
then consider P,
Q being
finite without_zero Subset of
NAT such that A2:
[:P,Q:] c= Indices M
and A3:
card P = card Q
and A4:
card P = RANK
and A5:
Det (EqSegm (M,P,Q)) <> 0. K
by Def4;
reconsider Sp =
Sgm P,
Sq =
Sgm Q as
Element of
RANK -tuples_on NAT by A3, A4;
A6:
rng Sp = P
by FINSEQ_1:def 14;
A7:
rng Sq = Q
by FINSEQ_1:def 14;
EqSegm (
M,
P,
Q) =
Segm (
M,
P,
Q)
by A3, Def3
.=
Segm (
M,
(Sgm P),
(Sgm Q))
;
hence
ex
rt1,
rt2 being
Element of
RANK -tuples_on NAT st
(
[:(rng rt1),(rng rt2):] c= Indices M &
Det (Segm (M,rt1,rt2)) <> 0. K )
by A2, A3, A4, A5, A6, A7;
for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK
let n be
Nat;
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANKlet nt1,
nt2 be
Element of
n -tuples_on NAT;
( [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K implies n <= RANK )
assume that A8:
[:(rng nt1),(rng nt2):] c= Indices M
and A9:
Det (Segm (M,nt1,nt2)) <> 0. K
;
n <= RANK
ex
P1,
P2 being
finite without_zero Subset of
NAT st
(
P1 = rng nt1 &
P2 = rng nt2 &
card P1 = card P2 &
card P1 = n &
Det (EqSegm (M,P1,P2)) <> 0. K )
by A8, A9, Th75;
hence
n <= RANK
by A1, A8, Def4;
verum
end;
assume that
A10:
ex rt1, rt2 being Element of RANK -tuples_on NAT st
( [:(rng rt1),(rng rt2):] c= Indices M & Det (Segm (M,rt1,rt2)) <> 0. K )
and
A11:
for n being Nat
for nt1, nt2 being Element of n -tuples_on NAT st [:(rng nt1),(rng nt2):] c= Indices M & Det (Segm (M,nt1,nt2)) <> 0. K holds
n <= RANK
; the_rank_of M = RANK
consider rt1, rt2 being Element of RANK -tuples_on NAT such that
A12:
[:(rng rt1),(rng rt2):] c= Indices M
and
A13:
Det (Segm (M,rt1,rt2)) <> 0. K
by A10;
consider P, Q being finite without_zero Subset of NAT such that
A14:
[:P,Q:] c= Indices M
and
A15:
card P = card Q
and
A16:
card P = the_rank_of M
and
A17:
Det (EqSegm (M,P,Q)) <> 0. K
by Def4;
ex P1, P2 being finite without_zero Subset of NAT st
( P1 = rng rt1 & P2 = rng rt2 & card P1 = card P2 & card P1 = RANK & Det (EqSegm (M,P1,P2)) <> 0. K )
by A12, A13, Th75;
then A18:
RANK <= card P
by A12, A16, Def4;
reconsider SP = Sgm P, SQ = Sgm Q as Element of (card P) -tuples_on NAT by A15;
A19:
rng SP = P
by FINSEQ_1:def 14;
A20:
rng SQ = Q
by FINSEQ_1:def 14;
EqSegm (M,P,Q) =
Segm (M,P,Q)
by A15, Def3
.=
Segm (M,(Sgm P),(Sgm Q))
;
then
card P <= RANK
by A11, A14, A15, A17, A19, A20;
hence
the_rank_of M = RANK
by A16, A18, XXREAL_0:1; verum