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 |
In mathematics, the spectral radius of a square matrix or a bounded linear operator is the supremum among the absolute values of the elements in its spectrum.
4
votes
Maximal eigenvalue of a correlation matrix with some entries fixed as zeros
The answer to your question is positive, in every dimension:
Let $A=I_n+T$ be symmetric positive semi-definite, with ${\rm diag}T=(0,\ldots,0)$ and $T$ tridiagonal. Then the largest eigenvalue $\rho( …
3
votes
Minimize spectral norm under diagonal similarity
The norm $A\mapsto s(A)$ has the flaw of not being unitarily invariant. However, if you think that every $P\in{\bf GL}_n({\mathbb R})$ can be factorized out $P=UDV$ where $U$ and $V$ are orthogonal an …
1
vote
A relation between norm and spectral radius for some matrix operators on Banach spaces $\ell...
If $A,A^T=\ell^p\rightarrow\ell^p$, then the adjoints $A^T,A$ map $\ell^{p'}$ into itself. By interpolation (Riesz-Thorin), they map $\ell^2$ into itself. It will be often the case that the spectrum o …
1
vote
How to derive the Euclidean norm of a matrix from its spectral radius
The correct statement is known as Householder Theorem: For every $\epsilon>0$, there exists a norm $\|\cdot\|$ over ${\mathbb C}^n$, such that the subordinated norm of $A$ is $\le\rho(A)+\epsilon$.
As …