TITLE: How to prove that the normalizer of diagonal matrices in $GL_n$ is the subgroup of generalized permutation matrices?
QUESTION [19 upvotes]: I'm trying to prove that de normalizer $N(T)$ of the subgroup $T\subset GL_n$ of diagonal matrices is the subgroup $P\in GL_n$ of generalized permutation matrices. I guess my biggest problem is that I don't really know how diagonal and permutation matrices (don't) commute. Because it is not true that $DM=MD$ when $D\in T$ and $M\in P$ since the permutation is either horizontal or vertical, but sometimes it seems like you can do something like it.
So far, I have proved that $P\subset N(T)$, in the following way. Let $M_\sigma\in P$. Then $M_\sigma=VS_\sigma$, with $V\in T$ and $S_\sigma$ a permutation matrix. So $M_\sigma DM^{-1}=VS_\sigma D S_\sigma^T V^{-1}$. Thus if we prove that $S_\sigma D S_\sigma^T$ is diagonal we are done. This is true since $S_\sigma D S_\sigma^T=(x_1 e_{\sigma(1)} \dots x_n e_{\sigma(n)}) (e_{\sigma^{-1}(1)} \dots e_{\sigma^{-1}(n)})=(x_{\sigma^{-1}(1)}e_1 \dots x_{\sigma^{-1}(n)}e_n)$, where $e_i$ are the standard basis vectors. Even this is hopelessly written out. I'm trying to find a way to see what the product $S_\sigma D S_\sigma^T$ is without writing it in vectors. 
For the other way around I don't really know what to do. I'm having a hard time rewriting matrix products in a useful way. Perhaps there is a way of proving this using something completely different? Maybe you can prove it using $N(T)/T\simeq S_n$, but this actually what I want to use my question for. When I just write what I know about a matrix $M\in N(T)$ I just get a big system of equations that isn't really handy.

REPLY [2 votes]: The set of eigenvectors common to all elements of $T$ is that of the nonzero multiples of the standard basis vectors. Any element of $N(T)$ must permute these common basis vectors among each other (if $P\in N(T)$ and $v$ is a common eigenvector of $T$, then so is $P\cdot v$). This means all columns of $P$ must have a single nonzero entry, and of course these entries have to be in distinct rows as well. Hence $N(T)$ is contained in the  set of generalised permutation matrices. The reverse inclusion follows from a simple computation to show that permutation matrices normalise $T$ (as of course do elements of $T$ itself). Or show this using the fact that $t\in T$ whenever all standard basis vectors are eigenvectors of $t$ (nearly the converse of the property used at the beginning).