Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
eigenvalues of matrices or operators
4
votes
Accepted
Similarity of two matrices
I will show that it is not possible for $\phi=\pi/2$, so it is certainly not for general $\phi$. (actually, I don't think that it is possible for any single $\phi$ except $0$ and $\pi$, by an analogou …
5
votes
Accepted
Eigenvalues invariant under 90° rotation
$$PXP^{-1}=Y$$
$$PYP^{-1}=X$$
$$QXQ^{-1}=X$$
$$QYQ^{-1}=-Y$$
from which it follows that
$$P(X+iY)P^{-1}=Y+iX=i(X-iY)$$
$$Q(X+iY)Q^{-1}=X-iY$$
This shows that $X+iY$ is conjugate to $i(X+iY)$, so its eigenvalues …
6
votes
On a matrix problem in the field $\mathbb F_2$
Some computation in sage yielded the following example, with $n=8$ and $P$ the cyclic permutation $(12345678)$:
$$M=\left(
\begin{array}{cc}
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 1 & 1 & …
20
votes
Accepted
Eigenvalue pattern
Thus the eigenvalues of $A$ are those of $\prod N_{\mu_i}$ plus those of $\overline{\prod N_{\mu_i}}$, which are their complex conjugates. … Moreover, since $N_\mu$ has determinant $1$, so does $\prod N_{\mu_i}$, so its two eigenvalues are inverses of each other. …