Question

Show $A = \{ u \in S^+(E) \textrm{ | } \forall x \in K, \langle x, u(x) \rangle \leq 1 \}$ is a compact set

score 75 · Answer 1 · 2026-03-25 17:40:41

75

Views

Show $A = \{ u \in S^+(E) \textrm{ | } \forall x \in K, \langle x, u(x) \rangle \leq 1 \}$ is a compact set

Published on 25 Mar 2026 - 17:40

score 1.1k · Answer 2 · 2026-03-25 15:59:02

1.1k

Views

Proof that compact self-adjoint operators have at least one non-zero eigenvector (using something analogous to min-max theorem)

Published on 25 Mar 2026 - 15:59

score 135 · Answer 3 · 2026-03-25 16:03:24

135

Views

Unique bounded operator constructed from sequence of given operators from orthogonal sum of Hilbert spaces to another Hilbert space.

Published on 25 Mar 2026 - 16:03

score 412 · Answer 4 · 2026-03-25 15:56:48

412

Views

"Square-normal" matrices are normal

Published on 25 Mar 2026 - 15:56

score 863 · Answer 5 · 2026-03-25 17:36:44

863

Views

How to prove an operator is invertible

Published on 25 Mar 2026 - 17:36

score 34 · Answer 6 · 2026-03-25 16:03:15

34

Views

Counter-example of a non-selfadjoint operator for which $ \left\| T \right\|= \sup_{x\in \mathcal{H},\left\| x \right\|=1} |(Tx,x)|$ does not hold.

Published on 25 Mar 2026 - 16:03

score 253 · Answer 7 · 2026-03-25 15:56:56

253

Views

Image self adjoint operator

Published on 25 Mar 2026 - 15:56

score 109 · Answer 8 · 2026-04-09 12:02:30

109

Views

Can changing the defined inner product change wheather a transformation is Normal or Self Adjoint?

Published on 09 Apr 2026 - 12:02

score 187 · Answer 9 · 2026-03-25 17:39:15

187

Views

When is the weak limit of self-adjoint invertible operators invertible?

Published on 25 Mar 2026 - 17:39

score 800 · Answer 10 · 2026-03-25 16:03:30

800

Views

Show that norm of normal operator equals spectral radius via $\| T T^* \| = \| T \|^2 = \| T \|^2$

Published on 25 Mar 2026 - 16:03

score 105 · Answer 11 · 2026-03-25 17:42:27

105

Views

Show that $T: L^2([0,1]) \to L^2([0,1])$, $f(x) \mapsto x \cdot f(x)$ is bounded.

Published on 25 Mar 2026 - 17:42

score 99 · Answer 12 · 2026-04-11 11:01:23

99

Views

Spectrum of product of self-adjoint operators contained in $\mathbb{R}$

Published on 11 Apr 2026 - 11:01

score 125 · Answer 13 · 2026-03-25 17:37:08

125

Views

Eigenvectors of Hermitian matrices over arbitrary fields

Published on 25 Mar 2026 - 17:37

score 188 · Answer 14 · 2026-04-17 14:00:18

188

Views

$\frac{(Tx,x)_H}{(x,x)_H}$ attains maximum if $T$ is compact and self-adjoint

Published on 17 Apr 2026 - 14:00

score 809 · Answer 15 · 2026-03-25 12:29:37

809

Views

How to show the Laplacian is a self-adjoint linear operator

Published on 25 Mar 2026 - 12:29

List Question

Show $A = \{ u \in S^+(E) \textrm{ | } \forall x \in K, \langle x, u(x) \rangle \leq 1 \}$ is a compact set

Proof that compact self-adjoint operators have at least one non-zero eigenvector (using something analogous to min-max theorem)

Unique bounded operator constructed from sequence of given operators from orthogonal sum of Hilbert spaces to another Hilbert space.

"Square-normal" matrices are normal

How to prove an operator is invertible

Counter-example of a non-selfadjoint operator for which $ \left\| T \right\|= \sup_{x\in \mathcal{H},\left\| x \right\|=1} |(Tx,x)|$ does not hold.

Image self adjoint operator

Can changing the defined inner product change wheather a transformation is Normal or Self Adjoint?

When is the weak limit of self-adjoint invertible operators invertible?

Show that norm of normal operator equals spectral radius via $\| T T^* \| = \| T \|^2 = \| T \|^2$

Show that $T: L^2([0,1]) \to L^2([0,1])$, $f(x) \mapsto x \cdot f(x)$ is bounded.

Spectrum of product of self-adjoint operators contained in $\mathbb{R}$

Eigenvectors of Hermitian matrices over arbitrary fields

$\frac{(Tx,x)_H}{(x,x)_H}$ attains maximum if $T$ is compact and self-adjoint

How to show the Laplacian is a self-adjoint linear operator

Trending Questions

Popular # Hahtags

Popular Questions