Question

An operator that is diagonalizable but not normal.

score 254 · Answer 1 · 2026-03-26 01:00:10

254

Views

An operator that is diagonalizable but not normal.

Published on 26 Mar 2026 - 1:00

score 65 · Answer 2 · 2026-03-26 04:34:13

65

Views

Proving that $0 \in \sigma (A)$ whenever $T$ is normal and $0 \in \sigma (T).$

Published on 26 Mar 2026 - 4:34

score 1.1k · Answer 3 · 2026-03-26 00:58:45

1.1k

Views

Proving that a Normal operator with real eigenvalues is self adjoint

Published on 26 Mar 2026 - 0:58

score 99 · Answer 4 · 2026-03-26 01:01:59

99

Views

Are normal operator with positive self-adjoint part boundedly invertible?

Published on 26 Mar 2026 - 1:01

score 479 · Answer 5 · 2026-03-26 01:04:36

479

Views

Adjoint of the direct sum of operators vs. direct sum of their adjoints

Published on 26 Mar 2026 - 1:04

score 39 · Answer 6 · 2026-03-26 01:04:39

39

Views

How to show that $\lim\limits_{j \to \infty} \lambda_j = 0$ and $P_j$'s are all finite rank projections?

Published on 26 Mar 2026 - 1:04

score 85 · Answer 7 · 2026-03-26 01:04:32

85

Views

$\text{ran}\,N$ is closed if and only if $0$ is no limit point of $\sigma(N)$, for a normal bounded operator $N$ on a hilbert space $H$

Published on 26 Mar 2026 - 1:04

score 149 · Answer 8 · 2026-03-26 01:07:54

149

Views

Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint

Published on 26 Mar 2026 - 1:07

score 60 · Answer 9 · 2026-03-26 01:04:43

60

Views

The isometry $S$ in polar decomposition of linear operator $L=S\sqrt{L^L}=SP$ has the same complex eigenvectors as $P$ iff $L^L=LL^*$.

Published on 26 Mar 2026 - 1:04

score 60 · Answer 10 · 2026-03-26 00:58:19

60

Views

Does $\mathbb{C}^n$ "contain" $\mathbb{R}^n$?

Published on 26 Mar 2026 - 0:58

score 47 · Answer 11 · 2026-03-26 01:06:39

47

Views

Spectral Theorem Multiplicationoperatorversion applied to a simple explicit example

Published on 26 Mar 2026 - 1:06

score 130 · Answer 12 · 2026-03-26 01:06:19

130

Views

If $T$ is normal, is ($T^2$ is compact $\Rightarrow T$ compact) true?

Published on 26 Mar 2026 - 1:06

score 937 · Answer 13 · 2026-03-26 01:04:40

937

Views

if $T$ is normal and a projection in a finite dimensional vector space $V$, then $T$ is an orthogonal projection

Published on 26 Mar 2026 - 1:04

score 88 · Answer 14 · 2026-03-26 01:03:27

88

Views

Spectral resolution of the identity of $S = U^*TU$

Published on 26 Mar 2026 - 1:03

score 281 · Answer 15 · 2026-03-26 01:03:03

281

Views

Normal operator $T$, $\sigma(T) \subset \{0,1\} \Rightarrow$ $T$ is an orthogonal projection

Published on 26 Mar 2026 - 1:03

List Question

An operator that is diagonalizable but not normal.

Proving that $0 \in \sigma (A)$ whenever $T$ is normal and $0 \in \sigma (T).$

Proving that a Normal operator with real eigenvalues is self adjoint

Are normal operator with positive self-adjoint part boundedly invertible?

Adjoint of the direct sum of operators vs. direct sum of their adjoints

How to show that $\lim\limits_{j \to \infty} \lambda_j = 0$ and $P_j$'s are all finite rank projections?

$\text{ran}\,N$ is closed if and only if $0$ is no limit point of $\sigma(N)$, for a normal bounded operator $N$ on a hilbert space $H$

Prove that if $A$ is a normal operator on an complex separable hilbert space $H$ and $A^{3} = A^{4}$ then $A$ is self adjoint

The isometry $S$ in polar decomposition of linear operator $L=S\sqrt{L^L}=SP$ has the same complex eigenvectors as $P$ iff $L^L=LL^*$.

Does $\mathbb{C}^n$ "contain" $\mathbb{R}^n$?

Spectral Theorem Multiplicationoperatorversion applied to a simple explicit example

If $T$ is normal, is ($T^2$ is compact $\Rightarrow T$ compact) true?

if $T$ is normal and a projection in a finite dimensional vector space $V$, then $T$ is an orthogonal projection

Spectral resolution of the identity of $S = U^*TU$

Normal operator $T$, $\sigma(T) \subset \{0,1\} \Rightarrow$ $T$ is an orthogonal projection

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