Metamath Proof Explorer


Theorem norm1exi

Description: A normalized vector exists in a subspace iff the subspace has a nonzero vector. (Contributed by NM, 9-Apr-2006) (New usage is discouraged.)

Ref Expression
Hypothesis norm1ex.1 ⊢ H ∈ S ℋ
Assertion norm1exi ⊢ ∃ x ∈ H x ≠ 0 ℎ ↔ ∃ y ∈ H norm ℎ ⁡ y = 1

Proof

Step Hyp Ref Expression
1 norm1ex.1 ⊢ H ∈ S ℋ
2 neeq1 ⊢ x = z → x ≠ 0 ℎ ↔ z ≠ 0 ℎ
3 2 cbvrexvw ⊢ ∃ x ∈ H x ≠ 0 ℎ ↔ ∃ z ∈ H z ≠ 0 ℎ
4 1 sheli ⊢ z ∈ H → z ∈ ℋ
5 normcl ⊢ z ∈ ℋ → norm ℎ ⁡ z ∈ ℝ
6 4 5 syl ⊢ z ∈ H → norm ℎ ⁡ z ∈ ℝ
7 6 adantr ⊢ z ∈ H ∧ z ≠ 0 ℎ → norm ℎ ⁡ z ∈ ℝ
8 normne0 ⊢ z ∈ ℋ → norm ℎ ⁡ z ≠ 0 ↔ z ≠ 0 ℎ
9 4 8 syl ⊢ z ∈ H → norm ℎ ⁡ z ≠ 0 ↔ z ≠ 0 ℎ
10 9 biimpar ⊢ z ∈ H ∧ z ≠ 0 ℎ → norm ℎ ⁡ z ≠ 0
11 7 10 rereccld ⊢ z ∈ H ∧ z ≠ 0 ℎ → 1 norm ℎ ⁡ z ∈ ℝ
12 11 recnd ⊢ z ∈ H ∧ z ≠ 0 ℎ → 1 norm ℎ ⁡ z ∈ ℂ
13 simpl ⊢ z ∈ H ∧ z ≠ 0 ℎ → z ∈ H
14 shmulcl ⊢ H ∈ S ℋ ∧ 1 norm ℎ ⁡ z ∈ ℂ ∧ z ∈ H → 1 norm ℎ ⁡ z ⋅ ℎ z ∈ H
15 1 14 mp3an1 ⊢ 1 norm ℎ ⁡ z ∈ ℂ ∧ z ∈ H → 1 norm ℎ ⁡ z ⋅ ℎ z ∈ H
16 12 13 15 syl2anc ⊢ z ∈ H ∧ z ≠ 0 ℎ → 1 norm ℎ ⁡ z ⋅ ℎ z ∈ H
17 norm1 ⊢ z ∈ ℋ ∧ z ≠ 0 ℎ → norm ℎ ⁡ 1 norm ℎ ⁡ z ⋅ ℎ z = 1
18 4 17 sylan ⊢ z ∈ H ∧ z ≠ 0 ℎ → norm ℎ ⁡ 1 norm ℎ ⁡ z ⋅ ℎ z = 1
19 fveqeq2 ⊢ y = 1 norm ℎ ⁡ z ⋅ ℎ z → norm ℎ ⁡ y = 1 ↔ norm ℎ ⁡ 1 norm ℎ ⁡ z ⋅ ℎ z = 1
20 19 rspcev ⊢ 1 norm ℎ ⁡ z ⋅ ℎ z ∈ H ∧ norm ℎ ⁡ 1 norm ℎ ⁡ z ⋅ ℎ z = 1 → ∃ y ∈ H norm ℎ ⁡ y = 1
21 16 18 20 syl2anc ⊢ z ∈ H ∧ z ≠ 0 ℎ → ∃ y ∈ H norm ℎ ⁡ y = 1
22 21 rexlimiva ⊢ ∃ z ∈ H z ≠ 0 ℎ → ∃ y ∈ H norm ℎ ⁡ y = 1
23 ax-1ne0 ⊢ 1 ≠ 0
24 23 neii ⊢ ¬ 1 = 0
25 eqeq1 ⊢ norm ℎ ⁡ y = 1 → norm ℎ ⁡ y = 0 ↔ 1 = 0
26 24 25 mtbiri ⊢ norm ℎ ⁡ y = 1 → ¬ norm ℎ ⁡ y = 0
27 1 sheli ⊢ y ∈ H → y ∈ ℋ
28 norm-i ⊢ y ∈ ℋ → norm ℎ ⁡ y = 0 ↔ y = 0 ℎ
29 27 28 syl ⊢ y ∈ H → norm ℎ ⁡ y = 0 ↔ y = 0 ℎ
30 29 necon3bbid ⊢ y ∈ H → ¬ norm ℎ ⁡ y = 0 ↔ y ≠ 0 ℎ
31 26 30 imbitrid ⊢ y ∈ H → norm ℎ ⁡ y = 1 → y ≠ 0 ℎ
32 31 reximia ⊢ ∃ y ∈ H norm ℎ ⁡ y = 1 → ∃ y ∈ H y ≠ 0 ℎ
33 neeq1 ⊢ y = z → y ≠ 0 ℎ ↔ z ≠ 0 ℎ
34 33 cbvrexvw ⊢ ∃ y ∈ H y ≠ 0 ℎ ↔ ∃ z ∈ H z ≠ 0 ℎ
35 32 34 sylib ⊢ ∃ y ∈ H norm ℎ ⁡ y = 1 → ∃ z ∈ H z ≠ 0 ℎ
36 22 35 impbii ⊢ ∃ z ∈ H z ≠ 0 ℎ ↔ ∃ y ∈ H norm ℎ ⁡ y = 1
37 3 36 bitri ⊢ ∃ x ∈ H x ≠ 0 ℎ ↔ ∃ y ∈ H norm ℎ ⁡ y = 1