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 |
11
votes
Kind of submultiplicativity of the Frobenius norm: $\|AB\|_F \leq \|A\|_2\|B\|_F$?
A simpler, more direct proof that requires no SVD: let $Y_j$ be the $j$th column of $Y$ and $Z_j$ that of $Z=XY$. Then,
$$\|Z\|_F^2 = \sum_j \|Z_j\|_2^2 = \sum_j \|XY_j\|_2^2 \leq \sum_j \|X\|_2^2\|Y_ …
3
votes
Can a perturbation of a matrix product always be represented as product of perturbations of ...
In numerical analysis lingo, you are more or less asking if matrix multiplication is backward stable. The answer seems to be no: see Section 3.5 of Higham, Accuracy and stability of numerical algorith …
2
votes
Accepted
Null Space Perturbations
What follows is just a trivial manipulation, but I do not think you can get better bounds than that for a generic perturbation. … leq \lVert \Delta A \rVert / \lVert A \rVert$ (up to first order in $\lVert \Delta A \rVert$, otherwise you have to take into account the additional term in the denominator depending on how large your perturbation …