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 linear-algebra
Search options not deleted
user 158036
Questions about the properties of vector spaces and linear transformations, including linear systems in general.
7
votes
Bounds on $\operatorname{sgn}(Au) - \operatorname{sgn}(Av)$ when $\|u-v\|_1 \leq \epsilon$
There is a straightforward bound.
Consider A to be the $\log(N)$-fold tensor product of $H= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$.
A is Unitary and Hermitian. In fact A …
- 221
3
votes
1
answer
107
views
Bounds on $\operatorname{sgn}(Au) - \operatorname{sgn}(Av)$ when $\|u-v\|_1 \leq \epsilon$
$\DeclareMathOperator\sgn{sgn}$Suppose A is a $N \times N$ Hermitian and unitary matrix, i.e., $A^{\dagger}=A$ and $A^{\dagger}A=I =AA^{\dagger}$. (Assume all entries are real.)
And let $u \in \{-1,1 …
- 221