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 6827
Questions about the properties of vector spaces and linear transformations, including linear systems in general.
7
votes
Accepted
ad (A^n) is a polynomial in ad A ?
I don't think it's true even when $A$ is diagonalisable.
Suppose that $A$ is a diagonal matrix, with $(i,i)$th entry $\lambda_i$; thus $A = \rm{diag}$$(\lambda_i)$. Then we can write $\rm{ad}(A) = \r …
- 3,702
29
votes
Accepted
When is $\ker AB = \ker A + \ker B$?
Since $\ker A = \ker A^2$, the map $\bar{A} : V/\ker A \to V/\ker A$ is injective. Since $V/\ker A$ is finite dimensional, this map is surjective. So for any $x \in V$ we can find $y \in V$ and $z \in …
- 3,702