List questions from eigenvalues-eigenvectors

Published on 25 Mar 2026 - 19:02

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$T$ and $U$ have a common eigenvector, where $T$ and $U$ are linear operators on odd dimensional vector space $V$ and $T^2 = U^2 = I$.

Published on 25 Mar 2026 - 9:38

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Distinct eigenvalues implies $A \in \mathbb{R}^{n \times n}$ is diagonalisable

Published on 25 Mar 2026 - 12:53

553

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Prove conditions for upper triangular matrix

Published on 16 Dec 2018 - 7:53

50

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Invariant subspace concept - generalization

Published on 25 Mar 2026 - 16:05

390

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I want to find largest eigenvalue using power iteration. Is it correct?

Published on 16 Dec 2018 - 17:19

64

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Determine all real numbers that can be eigenvalues of operator A.

Published on 26 Mar 2026 - 14:42

87

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If a square matrix $A$ has eigenvalue $1$, then $I-A$ is singular?

Published on 16 Dec 2018 - 22:34

154

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If we know the rank of a matrix r, can we assume that will have precisely r non-zero eigenvalues?

Published on 26 Mar 2026 - 7:45

915

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Absolute of all eigenvalues are always bounded by maximal singular value

Published on 26 Mar 2026 - 2:44

761

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Given this 4x4 matrix, is there an easier way to calculate the eigenvalues?

Published on 26 Mar 2026 - 7:42

164

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Sturm-Liouville eigenvalues have lower bound

Published on 27 Mar 2026 - 2:50

782

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Find locus of points by finding eigenvalues

Published on 17 Dec 2018 - 10:39

170

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Eigenvalues of Operator on $L^{2}[0, 1]$

Published on 01 Apr 2026 - 10:02

258

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Localization of eigenvalues

Published on 25 Mar 2026 - 17:35

List Question

Positive definite matrix if eigenvalue has positive and negative solutions

$T$ and $U$ have a common eigenvector, where $T$ and $U$ are linear operators on odd dimensional vector space $V$ and $T^2 = U^2 = I$.

Distinct eigenvalues implies $A \in \mathbb{R}^{n \times n}$ is diagonalisable

Prove conditions for upper triangular matrix

Invariant subspace concept - generalization

I want to find largest eigenvalue using power iteration. Is it correct?

Determine all real numbers that can be eigenvalues of operator A.

If a square matrix $A$ has eigenvalue $1$, then $I-A$ is singular?

If we know the rank of a matrix r, can we assume that will have precisely r non-zero eigenvalues?

Absolute of all eigenvalues are always bounded by maximal singular value

Given this 4x4 matrix, is there an easier way to calculate the eigenvalues?

Sturm-Liouville eigenvalues have lower bound

Find locus of points by finding eigenvalues

Eigenvalues of Operator on $L^{2}[0, 1]$

Localization of eigenvalues

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