Metamath Proof Explorer


Theorem lkrscss

Description: The kernel of a scalar product of a functional includes the kernel of the functional. (The inclusion is proper for the zero product and equality otherwise.) (Contributed by NM, 9-Oct-2014)

Ref Expression
Hypotheses lkrsc.v V = Base W
lkrsc.d D = Scalar W
lkrsc.k K = Base D
lkrsc.t · ˙ = D
lkrsc.f F = LFnl W
lkrsc.l L = LKer W
lkrsc.w φ W LVec
lkrsc.g φ G F
lkrsc.r φ R K
Assertion lkrscss φ L G L G · ˙ f V × R

Proof

Step Hyp Ref Expression
1 lkrsc.v V = Base W
2 lkrsc.d D = Scalar W
3 lkrsc.k K = Base D
4 lkrsc.t · ˙ = D
5 lkrsc.f F = LFnl W
6 lkrsc.l L = LKer W
7 lkrsc.w φ W LVec
8 lkrsc.g φ G F
9 lkrsc.r φ R K
10 lveclmod W LVec W LMod
11 7 10 syl φ W LMod
12 1 5 6 11 8 lkrssv φ L G V
13 eqid 0 D = 0 D
14 1 2 5 3 4 13 11 8 lfl0sc φ G · ˙ f V × 0 D = V × 0 D
15 14 fveq2d φ L G · ˙ f V × 0 D = L V × 0 D
16 eqid V × 0 D = V × 0 D
17 2 13 1 5 lfl0f W LMod V × 0 D F
18 2 13 1 5 6 lkr0f W LMod V × 0 D F L V × 0 D = V V × 0 D = V × 0 D
19 11 17 18 syl2anc2 φ L V × 0 D = V V × 0 D = V × 0 D
20 16 19 mpbiri φ L V × 0 D = V
21 15 20 eqtr2d φ V = L G · ˙ f V × 0 D
22 12 21 sseqtrd φ L G L G · ˙ f V × 0 D
23 22 adantr φ R = 0 D L G L G · ˙ f V × 0 D
24 sneq R = 0 D R = 0 D
25 24 xpeq2d R = 0 D V × R = V × 0 D
26 25 oveq2d R = 0 D G · ˙ f V × R = G · ˙ f V × 0 D
27 26 fveq2d R = 0 D L G · ˙ f V × R = L G · ˙ f V × 0 D
28 27 adantl φ R = 0 D L G · ˙ f V × R = L G · ˙ f V × 0 D
29 23 28 sseqtrrd φ R = 0 D L G L G · ˙ f V × R
30 7 adantr φ R 0 D W LVec
31 8 adantr φ R 0 D G F
32 9 adantr φ R 0 D R K
33 simpr φ R 0 D R 0 D
34 1 2 3 4 5 6 30 31 32 13 33 lkrsc φ R 0 D L G · ˙ f V × R = L G
35 eqimss2 L G · ˙ f V × R = L G L G L G · ˙ f V × R
36 34 35 syl φ R 0 D L G L G · ˙ f V × R
37 29 36 pm2.61dane φ L G L G · ˙ f V × R