TITLE: Commuting matrices and simultaneous diagonalizability
QUESTION [5 upvotes]: It is a known fact from linear algebra that if a set of matrices is pairwise commutable then they are simultaneously diagonalizable. A problem in the book I am currently studying asks to prove this claim using representation theory. Specifically
If $G = \{M_1,\ldots,M_k\}$ is an abelian subgroup of $\rm{GL}_d(\mathbb{C}).$ How can one use representation theory to show that the matrices in $G$ are simultaneously diagonalizable?

REPLY [4 votes]: This is more linear algebra than representation theory. One proof is to notice that all matrices $g$ in a finite subgroup $G$ of $\mathrm{GL}_n(\Bbb C)$ are diagonalisable, for if $m$ is the order of $G$, then $$g^m=\mathrm{id}.$$
That is, all elements of $G$ have a annihiliating polynomial with only simple roots, $X^m-1$. If $G$ is abelian, then te result you quoted shows that they are simultaneously diagonalisable.
If you want to use "representation theory", you could invoke the fact that all irreducible complex representations of an abelian group are one dimensional, and complete reducibility of complex representations of finite groups. But this is a corollary of the result above (the abelian part is.) The first paragraph of my answer is indeed the proof of this fact for abelian groups.