Question

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

score 74 · Answer 1 · 2019-12-10 11:56:54

74

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 10 Dec 2019 - 11:56

score 1.1k · Answer 2 · 2019-12-29 15:32:25

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 29 Dec 2019 - 15:32

score 134 · Answer 3 · 2020-01-03 18:55:11

134

Views

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

Published on 03 Jan 2020 - 18:55

score 411 · Answer 4 · 2020-01-05 00:20:14

411

Views

"Square-normal" matrices are normal

Published on 05 Jan 2020 - 0:20

score 862 · Answer 5 · 2020-01-15 18:11:29

862

Views

How to prove an operator is invertible

Published on 15 Jan 2020 - 18:11

score 33 · Answer 6 · 2020-01-17 11:28:42

33

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 17 Jan 2020 - 11:28

score 252 · Answer 7 · 2020-01-18 18:18:43

252

Views

Image self adjoint operator

Published on 18 Jan 2020 - 18:18

score 106 · Answer 8 · 2020-01-26 19:00:16

106

Views

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

Published on 26 Jan 2020 - 19:00

score 186 · Answer 9 · 2020-02-05 13:46:41

186

Views

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

Published on 05 Feb 2020 - 13:46

score 799 · Answer 10 · 2020-03-01 22:05:43

799

Views

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

Published on 01 Mar 2020 - 22:05

score 104 · Answer 11 · 2020-03-02 16:46:21

104

Views

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

Published on 02 Mar 2020 - 16:46

score 96 · Answer 12 · 2020-03-07 15:48:21

96

Views

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

Published on 07 Mar 2020 - 15:48

score 124 · Answer 13 · 2020-03-13 15:31:39

124

Views

Eigenvectors of Hermitian matrices over arbitrary fields

Published on 13 Mar 2020 - 15:31

score 185 · Answer 14 · 2020-03-15 11:26:57

185

Views

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

Published on 15 Mar 2020 - 11:26

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

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