Question

Are there any bounded linear operators on a Hilbert space such that its spectrum is exactly the unit circle?

score 244 · Answer 1 · 2026-03-27 05:58:49

244

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Are there any bounded linear operators on a Hilbert space such that its spectrum is exactly the unit circle?

Published on 27 Mar 2026 - 5:58

score 68 · Answer 2 · 2026-03-27 06:05:02

68

Views

Classify the spectrum of $T$ is $T$ is a bounded linear operator on a complex Hilbert space $H$ and $-T = T^*, T^2 = -I,$ and $T \not= \pm iI$

Published on 27 Mar 2026 - 6:05

score 49 · Answer 3 · 2026-03-25 07:48:31

49

Views

a surjective group homomorphism between $K_0$ group to an abelian group

Published on 25 Mar 2026 - 7:48

score 96 · Answer 4 · 2026-03-25 03:01:24

96

Views

Interpolation type theorem for multiplication operator on measure space. Is it trivial?

Published on 25 Mar 2026 - 3:01

score 40 · Answer 5 · 2026-03-29 18:31:32

40

Views

Show that if $||T||$ is an eigenvalue for $T$, it is also an eigenvalue for $T^*$

Published on 29 Mar 2026 - 18:31

score 86 · Answer 6 · 2026-03-29 20:54:59

86

Views

Does The Riemann-Siegel function define a unitary operator In Complex Hilbert space?

Published on 29 Mar 2026 - 20:54

score 64 · Answer 7 · 2026-03-28 00:27:37

64

Views

Increasing sequence of positive operators

Published on 28 Mar 2026 - 0:27

score 69 · Answer 8 · 2026-03-27 23:41:02

69

Views

Why are operators often written in 3s?

Published on 27 Mar 2026 - 23:41

score 30 · Answer 9 · 2026-03-26 01:10:56

30

Views

Is an operator-valued function of bounded variation if and only if it is the difference of nondecreasing functions?

Published on 26 Mar 2026 - 1:10

score 61 · Answer 10 · 2026-03-25 09:28:44

61

Views

Show that the resolvent set is a null set of the spectral measure associated with a self-adjoint operator

Published on 25 Mar 2026 - 9:28

score 463 · Answer 11 · 2019-08-13 18:57:15

463

Views

Norm of a block of matrix operator

Published on 13 Aug 2019 - 18:57

score 28 · Answer 12 · 2019-08-13 22:46:52

28

Views

Second method for finding matrix operator $f$ in base $\beta_2$

Published on 13 Aug 2019 - 22:46

score 492 · Answer 13 · 2026-03-27 05:59:37

492

Views

If $K$ is a compact, self-adjoint linear operator on a separable Hilbert space with closed image, prove that the image of $K$ is finite dimensional

Published on 27 Mar 2026 - 5:59

score 74 · Answer 14 · 2026-03-27 06:04:58

74

Views

If $E$ is the spectral measure coresponding to a self-adjoint operator, show $\int1+2\sum_{i=1}^n\frac{n-i}nλ^idE(λ)f→\int\frac{1+λ}{1-λ}dE(λ)f$

Published on 27 Mar 2026 - 6:04

score 83 · Answer 15 · 2026-03-28 17:39:25

83

Views

Alternative proofs invertible matrix/operator $T$ has square root over complex field using polar decomposition

Published on 28 Mar 2026 - 17:39

List Question

Are there any bounded linear operators on a Hilbert space such that its spectrum is exactly the unit circle?

Classify the spectrum of $T$ is $T$ is a bounded linear operator on a complex Hilbert space $H$ and $-T = T^*, T^2 = -I,$ and $T \not= \pm iI$

a surjective group homomorphism between $K_0$ group to an abelian group

Interpolation type theorem for multiplication operator on measure space. Is it trivial?

Show that if $||T||$ is an eigenvalue for $T$, it is also an eigenvalue for $T^*$

Does The Riemann-Siegel function define a unitary operator In Complex Hilbert space?

Increasing sequence of positive operators

Why are operators often written in 3s?

Is an operator-valued function of bounded variation if and only if it is the difference of nondecreasing functions?

Show that the resolvent set is a null set of the spectral measure associated with a self-adjoint operator

Norm of a block of matrix operator

Second method for finding matrix operator $f$ in base $\beta_2$

If $K$ is a compact, self-adjoint linear operator on a separable Hilbert space with closed image, prove that the image of $K$ is finite dimensional

If $E$ is the spectral measure coresponding to a self-adjoint operator, show $\int1+2\sum_{i=1}^n\frac{n-i}nλ^idE(λ)f→\int\frac{1+λ}{1-λ}dE(λ)f$

Alternative proofs invertible matrix/operator $T$ has square root over complex field using polar decomposition

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