Prove two commute, diagonalizable matrices have the some diagonalizing matrix Let $\ M \in \mathbb M_{n\times n} $ be a finite group of diagonalizable matrices that each two are commuting ($\ AB = BA $ ), I need to prove that there exists a $\ P $ matrix that will diagonalize any matrix in $\ M $
I can't intuitively understand why it is true, and can't really think of a way to prove it. If I understand correctly, according to this paper it is just called Schur lemma?
I've found many questions about opposite direction (if they diagonalizable then they commute)
 A: This is a classical result, so you can find a proof in any course on reduction of endomorphism. See for instance theorem 5.11 in:
https://pub.math.leidenuniv.nl/~luijkrmvan/linalg2/2018/LinAlg2-2018.pdf
(which is the first one I found).
NB1 : If you are looking at a finite subgroup $G$ of $Gl_n(\mathbb C)$, the matrices in $G$ are automatically diagonalizable, since they have finite order. See Commuting matrices and simultaneous diagonalizability
NB2 : Over $\mathbb C$, you can see it as a consequence of Schur's Lemma, but in a contorted way. Namely, Schur's Lemma says that over an algebraically closed field, the only endomorphisms of an irreducible representation $V$ of a group $G$ are homotheties. If your group is abelian, then the action $g \cdot -$ of an element $g \in G$ gives an endomorphism of $V$, so $g$ acts as an homothety, hence any subspace of $V$ is stable by any $g$, and $V$ must be of dimension $1$ (else it would not be irreducible). If your group is finite, any representation is a direct sum of irreducible representations. So, if $G$ is abelian and finite, any representation is a direct sum of stable lines, and a basis adapted to this decomposition is a co-diagonalization basis for the action of $G$.
