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 |
Questions about the properties of vector spaces and linear transformations, including linear systems in general.
5
votes
Submultiplicative matrix norm: Max Norm
The Hilbert-Schmidt norm, $||A||_F= (\sum_{i,j=1}^n a_{i,j}^2)^{1/2}$ is clearly always larger than $||A||_{max}$ and is also submultiplicative.
Hence, $||AB||_{max} \leq ||AB||_F \leq ||A||_F ||B||_ …
2
votes
1
answer
354
views
The operator equation $AB = \lambda BA$ for self-adjoint operators
Suppose that $A$ and $B$ are self-adjoint bounded linear operators on a Hilbert space and $\lambda \in \mathbb{C}$. It turns out that if $\lambda \notin \{-1, 1\}$ then $AB=\lambda BA \implies AB = B …
1
vote
Topics for a matrix analysis course
I wrote my dissertation on a problem in matrix analysis and I found that I had to read from several different sources to understand the material. I don't know which is the best book on the subject bu …
11
votes
Condition for two matrices to share at least one eigenvector?
This may be a partial solution to your problem:
I claim that if there exists a shared eigenvector, $x$ of $A$ and $B$ with common eigenvalue of $1$ then $\det(AB - BA) = \det[A,B] = 0$.
Proof:
Supp …
0
votes
Norm of triangular truncation operator on rank deficient matrices
Fix $r \geq 1$ and let $A_n$ be the $n \times n$ identity matrix with the bottom $n-r$ rows replaced with rows of zeros. Then $||A_n|| = 1$ for all $n$ and $||T_n \circ A_n|| = ||A_n||$ so we have th …