Question

Can the inner product of two negative vectors be negative?

score 286 · Answer 1 · 2026-04-09 20:08:27

286

Views

Can the inner product of two negative vectors be negative?

Published on 09 Apr 2026 - 20:08

score 980 · Answer 2 · 2026-04-11 23:09:40

980

Views

What's the purpose of defining the inner product?

Published on 11 Apr 2026 - 23:09

score 518 · Answer 3 · 2026-05-10 13:31:31

518

Views

Show that $U^\perp=\{0\}$.

Published on 10 May 2026 - 13:31

score 80 · Answer 4 · 2026-05-10 20:00:30

80

Views

Does this function correspond to some inner product?

Published on 10 May 2026 - 20:00

score 753 · Answer 5 · 2026-05-10 20:02:56

753

Views

Looking for some geometric intuition for half spaces in $\mathbb{R}^{d}$

Published on 10 May 2026 - 20:02

score 610 · Answer 6 · 2026-05-14 01:12:30

610

Views

General Cauchy-Schwarz inequality proof using discriminant

Published on 14 May 2026 - 1:12

score 231 · Answer 7 · 2026-05-10 15:50:55

231

Views

Positive-Definite linear operator as an inner product

Published on 10 May 2026 - 15:50

score 591 · Answer 8 · 2026-04-13 08:39:32

591

Views

Why is the inner product always real?

Published on 13 Apr 2026 - 8:39

score 839 · Answer 9 · 2026-04-16 02:07:42

839

Views

Show $\langle p,q\rangle$ is an inner product on a vector space

Published on 16 Apr 2026 - 2:07

score 36 · Answer 10 · 2026-05-10 21:45:18

36

Views

Concluding orthogonality

Published on 10 May 2026 - 21:45

score 326 · Answer 11 · 2026-04-15 03:37:15

326

Views

$⟨v, w⟩_1 = 0$ if and only if $⟨v, w⟩_2 = 0$. Prove that $⟨v, w⟩_1 = c⟨v, w⟩_2$, infinite-dimensional case?

Published on 15 Apr 2026 - 3:37

score 332 · Answer 12 · 2026-04-17 14:48:25

332

Views

Finding $g_t$ such that $(f|g)=f(t)$

Published on 17 Apr 2026 - 14:48

score 367 · Answer 13 · 2026-05-10 14:59:09

367

Views

Find orthonormal basis given inner product.

Published on 10 May 2026 - 14:59

score 39 · Answer 14 · 2026-04-09 23:44:54

39

Views

Show that if $A$ is a bounded linear operator on a Hilbert space then $||Ax||^2 \le ||A||\langle Ax,x\rangle$

Published on 09 Apr 2026 - 23:44

score 190 · Answer 15 · 2026-05-10 16:53:08

190

Views

Prove it is possible $Tv =\lambda v$ while v is not an eigenvector of T* with eigenvalue $\overline{\lambda}$.

Published on 10 May 2026 - 16:53

List Question

Can the inner product of two negative vectors be negative?

What's the purpose of defining the inner product?

Show that $U^\perp=\{0\}$.

Does this function correspond to some inner product?

Looking for some geometric intuition for half spaces in $\mathbb{R}^{d}$

General Cauchy-Schwarz inequality proof using discriminant

Positive-Definite linear operator as an inner product

Why is the inner product always real?

Show $\langle p,q\rangle$ is an inner product on a vector space

Concluding orthogonality

$⟨v, w⟩_1 = 0$ if and only if $⟨v, w⟩_2 = 0$. Prove that $⟨v, w⟩_1 = c⟨v, w⟩_2$, infinite-dimensional case?

Finding $g_t$ such that $(f|g)=f(t)$

Find orthonormal basis given inner product.

Show that if $A$ is a bounded linear operator on a Hilbert space then $||Ax||^2 \le ||A||\langle Ax,x\rangle$

Prove it is possible $Tv =\lambda v$ while v is not an eigenvector of T* with eigenvalue $\overline{\lambda}$.

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