Question

Why are $-1,1$ the only diagonalizable eigenvalues for orthogonal operators?

score 81 · Answer 1 · 2026-03-28 13:11:06

81

Views

Why are $-1,1$ the only diagonalizable eigenvalues for orthogonal operators?

Published on 28 Mar 2026 - 13:11

score 99 · Answer 2 · 2026-02-23 12:04:32

99

Views

Vector space$V$ is direct sum of eigenspaces of $\phi$ when $\phi $ is diagonalisable

Published on 23 Feb 2026 - 12:04

score 48 · Answer 3 · 2023-04-28 20:15:37

48

Views

If eigenspaces $V_{\lambda} (\phi )= V_{\lambda} (\phi ‘)$ then $ \phi = \phi ‘$

Published on 28 Apr 2023 - 20:15

score 255 · Answer 4 · 2026-03-28 13:11:11

255

Views

Visualising geometrically $\lambda=\pm 1$ eigenvalues for orthogonal transformations

Published on 28 Mar 2026 - 13:11

score 57 · Answer 5 · 2023-05-02 03:04:14

57

Views

Diagonalization method on triangular matrix to compute large power

Published on 02 May 2023 - 3:04

score 46 · Answer 6 · 2023-05-08 21:15:31

46

Views

Matrix antidiagonalization

Published on 08 May 2023 - 21:15

score 46 · Answer 7 · 2023-05-10 00:42:03

46

Views

Explicit diagonalization of $(e^{i \alpha (k-l)})_{1 \leq k,l \leq N}$.

Published on 10 May 2023 - 0:42

score 65 · Answer 8 · 2026-03-25 05:59:53

65

Views

Diagonalize block matrix containing circulant and sparse matrices

Published on 25 Mar 2026 - 5:59

score 52 · Answer 9 · 2026-03-28 07:37:14

52

Views

Prove $T$ is diagonalisable if $T^*=\alpha T$ for some $\alpha\in\mathbb{C}$

Published on 28 Mar 2026 - 7:37

score 51 · Answer 10 · 2026-03-28 07:38:36

51

Views

Prove there is an orthogonal basis for $V$ of eigenvectors for both $T$ and adjoint $T^*$

Published on 28 Mar 2026 - 7:38

score 43 · Answer 11 · 2023-05-13 11:34:10

43

Views

Show that if $Sp(f)=\{\lambda_1,...,\lambda_n\}$ then $Sp(Q(f))=\{Q(\lambda_1), ..., Q(\lambda_n)\}$

Published on 13 May 2023 - 11:34

score 44 · Answer 12 · 2023-05-14 13:33:21

44

Views

Geometrically, what is unique about the mapping of an orthogonally diagonizable matrix (versus non-orthogonal)?

Published on 14 May 2023 - 13:33

score 59 · Answer 13 · 2023-05-14 20:41:29

59

Views

Solving a differential equation with given eigenvalues and eigenvectors

Published on 14 May 2023 - 20:41

score 69 · Answer 14 · 2023-05-21 10:19:08

69

Views

Diagonalizing matrices over $\mathbb{Z}/ p^k$

Published on 21 May 2023 - 10:19

score 132 · Answer 15 · 2026-03-30 23:04:37

132

Views

Proving that $A=A(B−I)$ and $B=B(A−I)$ implies $A$ is diagonalizable.

Published on 30 Mar 2026 - 23:04

List Question

Why are $-1,1$ the only diagonalizable eigenvalues for orthogonal operators?

Vector space$V$ is direct sum of eigenspaces of $\phi$ when $\phi $ is diagonalisable

If eigenspaces $V_{\lambda} (\phi )= V_{\lambda} (\phi ‘)$ then $ \phi = \phi ‘$

Visualising geometrically $\lambda=\pm 1$ eigenvalues for orthogonal transformations

Diagonalization method on triangular matrix to compute large power

Matrix antidiagonalization

Explicit diagonalization of $(e^{i \alpha (k-l)})_{1 \leq k,l \leq N}$.

Diagonalize block matrix containing circulant and sparse matrices

Prove $T$ is diagonalisable if $T^*=\alpha T$ for some $\alpha\in\mathbb{C}$

Prove there is an orthogonal basis for $V$ of eigenvectors for both $T$ and adjoint $T^*$

Show that if $Sp(f)=\{\lambda_1,...,\lambda_n\}$ then $Sp(Q(f))=\{Q(\lambda_1), ..., Q(\lambda_n)\}$

Geometrically, what is unique about the mapping of an orthogonally diagonizable matrix (versus non-orthogonal)?

Solving a differential equation with given eigenvalues and eigenvectors

Diagonalizing matrices over $\mathbb{Z}/ p^k$

Proving that $A=A(B−I)$ and $B=B(A−I)$ implies $A$ is diagonalizable.

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