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 19184
Questions about the properties of vector spaces and linear transformations, including linear systems in general.
4
votes
The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
Let me simplify a bit (if I may) a nice argument by Matthew Emerton, by omitting the part with $p$-reductions. We start as above: let $g \in\mathrm{GL}(n,\mathbf Q)$ normalize $\mathrm{GL}(n,\mathbf Z …
- 303
15
votes
3
answers
1k
views
The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is,
$$
N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z) …
- 303