Metamath Proof Explorer


Theorem hdmapinvlem1

Description: Line 27 in Baer p. 110. We use C for Baer's u. Our unit vector E has the required properties for his w by hdmapevec2 . Our ( ( SE )C ) means the inner product <. C , E >. i.e. his f(u,w) (note argument reversal). (Contributed by NM, 11-Jun-2015)

Ref Expression
Hypotheses hdmapinvlem1.h ⊢ H = LHyp ⁡ K
hdmapinvlem1.e ⊢ E = I ↾ Base K I ↾ LTrn ⁡ K ⁡ W
hdmapinvlem1.o ⊢ O = ocH ⁡ K ⁡ W
hdmapinvlem1.u ⊢ U = DVecH ⁡ K ⁡ W
hdmapinvlem1.v ⊢ V = Base U
hdmapinvlem1.r ⊢ R = Scalar ⁡ U
hdmapinvlem1.b ⊢ B = Base R
hdmapinvlem1.t ⊢ · ˙ = ⋅ R
hdmapinvlem1.z ⊢ 0 ˙ = 0 R
hdmapinvlem1.s ⊢ S = HDMap ⁡ K ⁡ W
hdmapinvlem1.k ⊢ φ → K ∈ HL ∧ W ∈ H
hdmapinvlem1.c ⊢ φ → C ∈ O ⁡ E
Assertion hdmapinvlem1 ⊢ φ → S ⁡ E ⁡ C = 0 ˙

Proof

Step Hyp Ref Expression
1 hdmapinvlem1.h ⊢ H = LHyp ⁡ K
2 hdmapinvlem1.e ⊢ E = I ↾ Base K I ↾ LTrn ⁡ K ⁡ W
3 hdmapinvlem1.o ⊢ O = ocH ⁡ K ⁡ W
4 hdmapinvlem1.u ⊢ U = DVecH ⁡ K ⁡ W
5 hdmapinvlem1.v ⊢ V = Base U
6 hdmapinvlem1.r ⊢ R = Scalar ⁡ U
7 hdmapinvlem1.b ⊢ B = Base R
8 hdmapinvlem1.t ⊢ · ˙ = ⋅ R
9 hdmapinvlem1.z ⊢ 0 ˙ = 0 R
10 hdmapinvlem1.s ⊢ S = HDMap ⁡ K ⁡ W
11 hdmapinvlem1.k ⊢ φ → K ∈ HL ∧ W ∈ H
12 hdmapinvlem1.c ⊢ φ → C ∈ O ⁡ E
13 eqid ⊢ LFnl ⁡ U = LFnl ⁡ U
14 eqid ⊢ LKer ⁡ U = LKer ⁡ U
15 eqid ⊢ Base K = Base K
16 eqid ⊢ LTrn ⁡ K ⁡ W = LTrn ⁡ K ⁡ W
17 eqid ⊢ 0 U = 0 U
18 1 15 16 4 5 17 2 11 dvheveccl ⊢ φ → E ∈ V ∖ 0 U
19 18 eldifad ⊢ φ → E ∈ V
20 1 3 4 5 13 14 10 11 19 hdmaplkr ⊢ φ → LKer ⁡ U ⁡ S ⁡ E = O ⁡ E
21 12 20 eleqtrrd ⊢ φ → C ∈ LKer ⁡ U ⁡ S ⁡ E
22 1 4 11 dvhlmod ⊢ φ → U ∈ LMod
23 eqid ⊢ LCDual ⁡ K ⁡ W = LCDual ⁡ K ⁡ W
24 eqid ⊢ Base LCDual ⁡ K ⁡ W = Base LCDual ⁡ K ⁡ W
25 1 4 5 23 24 10 11 19 hdmapcl ⊢ φ → S ⁡ E ∈ Base LCDual ⁡ K ⁡ W
26 1 23 24 4 13 11 25 lcdvbaselfl ⊢ φ → S ⁡ E ∈ LFnl ⁡ U
27 19 snssd ⊢ φ → E ⊆ V
28 1 4 5 3 dochssv ⊢ K ∈ HL ∧ W ∈ H ∧ E ⊆ V → O ⁡ E ⊆ V
29 11 27 28 syl2anc ⊢ φ → O ⁡ E ⊆ V
30 29 12 sseldd ⊢ φ → C ∈ V
31 5 6 9 13 14 22 26 30 ellkr2 ⊢ φ → C ∈ LKer ⁡ U ⁡ S ⁡ E ↔ S ⁡ E ⁡ C = 0 ˙
32 21 31 mpbid ⊢ φ → S ⁡ E ⁡ C = 0 ˙