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 eigenvector
Search options not deleted
user 164237
0
votes
0
answers
79
views
Eigendecomposition of $A=I+BDB^H$
Suppose that we have $$A = I_m + BDB^H$$ where matrix $A$ is $m \times m$, matrix $B$ is $m \times k$, $BB^H \neq I_m$ and $D$ is a $k \times k$ diagonal matrix. Can we obtain the eigendecomposition …
- 13