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 matrices
Search options not deleted
user 8472
Questions where the notion matrix has an important or crucial role (for the latter, note the tag matrix-theory for potential use). Matrices appear in various parts of mathematics, and this tag is typically combined with other tags to make the general subject clear, such as an appropriate top-level tag ra.rings-and-algebras, co.combinatorics, etc. and other tags that might be applicable. There are also several more specialized tags concerning matrices.
2
votes
is there any efficient way to compute the follow matrix equations easily
The
technique I described avoids some of the problems which might arise in linear system solution,
i.e. ill-conditioning of the coefficient matrices. …
- 391
4
votes
Determinants of "almost identity" matrices.
in fedja's comment: for $s \in [0, 1]$ the
matrix $sB$ exhibits all the properties which have been shown to hold for $B$, so
$s \to (I - sB)$ is a continuous path from $I$ to $A$ through nonsingular matrices … Finally, I guess you can see from what has been
posted here that such matrices are quite well known. …
- 391