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 | |
Results tagged with matrix-analysis
Search options answers only
not deleted
user 766
The study of the properties of real and complex matrices that are more close to analysis and operator theory. For instance: the properties of positive definite matrices, matrix inequalities, perturbation analysis, matrix functions, inequalities between eigenvectors and singular values, majorization.
9
votes
Accepted
A question about the paper "The Condition Number of a Randomly Perturbed Matrix"
One does not need to have $n^{-B-3/2}/2$ to be equal to $0.1$, it is enough for it to be less than or equal to $0.1$, which is certainly the case for $n$ large enough.
Thanks for pointing out this ty …
- 114k
11
votes
Accepted
Product $PVPVP$ is elementwise nonnegative?
First, we repeat the arguments from this stackexchange answer. $P^{-1}$ is an $M$-matrix, and can thus be written as $s(I-A)$ for some positive $s$ and some $A$ with non-negative entries. As $P^{-1} …
CommunityBot
- 1
29
votes
Accepted
A curious determinantal inequality
Let $C := A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2}$; this is a positive semi-definite matrix with the same trace as $(A+B)^2$. We show that the eigenvalues of $C$ are majorised by the eigenvalue …
- 105k
9
votes
A curious determinantal inequality
EDIT: the argument below is not correct, but I am leaving it here in case it is of use in locating a better solution.
By a limiting argument we may assume that $C := A+B$ is invertible. If we write
…
- 114k
5
votes
Accepted
Is the p-norm of a matrix strictly log-convex?
The answer is "no".
First, a trivial counterexample: let $A_n$ be the $n \times n$ matrix with all $1$s on the first row and zeroes elsewhere, then $\frac{1}{p} \mapsto \|A_n\|_p = n^{1/p}$ is log-co …
- 114k