TITLE: Simultaneous diagonalisable matrices
QUESTION [6 upvotes]: I am well aware that there are already several questions and posts regarding the following topic. However, I could not find any answer to the following problem in Bruce Sagan's book The Symmetric Group:
Let $\left\{X_1,X_2,...X_n\right\} \subseteq GL_d$ be a subgroup of commuting matrices. Show that these matrices are simultaneously diagonalisable using representation theory.

Two $n$ x $n$ matrices $A, B$ are said to be simultaneously
  diagonalizable if there is an invertible matrix $S$ such that both
  $S^{-1}AS$ and $S^{-1}BS$ are diagonal matrices.

I have already read the question Commuting matrices and simultaneous diagonalizability and all the other posts I could find, but I think that a proof should involve Schur's Lemma, as is hinted by Proposition 2.8 on page 5 in these lecture notes.
Do you know how to find an answer to this problem?
I am sorry that I could not yet find an answer on how to prove this!
Thank you very much for your help!

REPLY [2 votes]: I'm not an expert in representation theory but believe that over complex numbers, the main idea goes as follows. The matrices form a commutative subgroup $G$ of $GL$. Any irreducible subrepresentation of the defining representation $\mathbb{C}^d$ of $G$ is one-dimensional (the action of any $X_i$ is $G$-invariant, so it acts as a scalar by Schur theorem), so an irreducible subrepresentation generates a common eigenvector to all your matrices. Now $G$ is finite, hence the representation $\mathbb{C}^d$ is completely reducible and you can find a $G$-invariant complement (and, by induction, further eigenvectors common to all matrices).
Note, however, that the result would not be true if $G=\{X_1, \ldots, X_n\}$ were not a group (the counterexample being $X_i=\begin{pmatrix}1 & i\\ 0 & 1\end{pmatrix}$ for $i=1,2$).