Question

Infinite-dimensional inner product space: if $A \geq 0$ and if $\langle Ax, x\rangle = 0$ for some $x$, then $Ax = 0$.

score 317 · Answer 1 · 2020-06-23 01:41:27

317

Views

Infinite-dimensional inner product space: if $A \geq 0$ and if $\langle Ax, x\rangle = 0$ for some $x$, then $Ax = 0$.

Published on 23 Jun 2020 - 1:41

score 186 · Answer 2 · 2020-06-23 15:45:45

186

Views

Show $(A-\lambda)^{-1}=-\int_0^\infty e^{-\lambda t}e^{tA}dt$ using spectral theorem

Published on 23 Jun 2020 - 15:45

score 103 · Answer 3 · 2026-03-25 12:15:20

103

Views

Limit with unbounded operator and spectral measure

Published on 25 Mar 2026 - 12:15

score 288 · Answer 4 · 2020-07-02 05:27:39

288

Views

Product of operator with its adjoint is self-adjoint

Published on 02 Jul 2020 - 5:27

score 69 · Answer 5 · 2020-07-06 00:29:06

69

Views

Define the map $T(f)=a(x)f(x)$ where $a(x) \in C[-\pi,\pi]$ and $f \in L^2$, show that $T$ is not compact unless $a=0$

Published on 06 Jul 2020 - 0:29

score 169 · Answer 6 · 2020-07-10 12:13:52

169

Views

If $A$ is positive and invertible, and $B$ is Hermitian, then $A+iB$ is invertible.

Published on 10 Jul 2020 - 12:13

score 84 · Answer 7 · 2020-07-12 01:25:45

84

Views

Self adjoint operators leaves invariant some unidimensional subspace

Published on 12 Jul 2020 - 1:25

score 93 · Answer 8 · 2020-07-13 08:25:07

93

Views

If $A \leq B$ for a positive-definite operator $A$ in finite dimensions, then $B^{-1} \leq A^{-1}$

Published on 13 Jul 2020 - 8:25

score 424 · Answer 9 · 2020-07-20 20:52:16

424

Views

The geometry meaning of Unitary matrix/operator

Published on 20 Jul 2020 - 20:52

score 1.5k · Answer 10 · 2026-03-25 17:36:18

1.5k

Views

Is every unitary operator a self-adjoint one?

Published on 25 Mar 2026 - 17:36

score 40 · Answer 11 · 2020-07-28 09:43:59

40

Views

How to interpret $P_{V^{\perp}}m$, where $P_{V^{\perp}}$ is an orthogonal projection and $m$ is a matrix?

Published on 28 Jul 2020 - 9:43

score 111 · Answer 12 · 2020-07-29 16:44:50

111

Views

Showing the operator in the Maxwell eigenproblem is self-adjoint

Published on 29 Jul 2020 - 16:44

score 32 · Answer 13 · 2026-03-19 23:51:23

32

Views

$A^*+A\geq0$ if and only if $A\geq0$

Published on 19 Mar 2026 - 23:51

score 34 · Answer 14 · 2026-03-25 17:22:51

34

Views

Differential of the Trace for functions applied to unbounded self-adjoint operators

Published on 25 Mar 2026 - 17:22

score 171 · Answer 15 · 2026-02-23 06:38:04

171

Views

Strongly continuous one-parameter unitary group and the form domain of the generator

Published on 23 Feb 2026 - 6:38

List Question

Infinite-dimensional inner product space: if $A \geq 0$ and if $\langle Ax, x\rangle = 0$ for some $x$, then $Ax = 0$.

Show $(A-\lambda)^{-1}=-\int_0^\infty e^{-\lambda t}e^{tA}dt$ using spectral theorem

Limit with unbounded operator and spectral measure

Product of operator with its adjoint is self-adjoint

Define the map $T(f)=a(x)f(x)$ where $a(x) \in C[-\pi,\pi]$ and $f \in L^2$, show that $T$ is not compact unless $a=0$

If $A$ is positive and invertible, and $B$ is Hermitian, then $A+iB$ is invertible.

Self adjoint operators leaves invariant some unidimensional subspace

If $A \leq B$ for a positive-definite operator $A$ in finite dimensions, then $B^{-1} \leq A^{-1}$

The geometry meaning of Unitary matrix/operator

Is every unitary operator a self-adjoint one?

How to interpret $P_{V^{\perp}}m$, where $P_{V^{\perp}}$ is an orthogonal projection and $m$ is a matrix?

Showing the operator in the Maxwell eigenproblem is self-adjoint

$A^*+A\geq0$ if and only if $A\geq0$

Differential of the Trace for functions applied to unbounded self-adjoint operators

Strongly continuous one-parameter unitary group and the form domain of the generator

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