Let $A,B \in M_n (\mathbb{C})$ be normal matrices such that $AB = BA$. Prove that these matrices can be simultaneously unitarily diagonalized.
A matrix $U$ is unitary if and only if $U^{*} = U^{-1}$. Simultaneous diagonalization by a unitary matrix therefore means there exists a unitary $U$ such that $U^{-1} AU = U^{*} AU$ is diagonal and $U^{-1} BU = U^{*} BU$ is diagonal. I believe I need to first show that $A$ and $B$ have the same eigenvalues; if $u$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$ ABu = BAu = \lambda Bu, $$ which tells me that $Bu$ is an eigenvector of $A$ with the same eigenvalue $\lambda$ provided that $Bu \neq 0$, but that doesn't tell me anything about the eigenvectors of $B$.
Any help would be appreciated.
Recall that the eigenspaces of $A$ are the same as the eigenspaces of $A^*$ (by normality). Then show that these $A/A^*$-eigenspaces are both $B$ and $B^*$-invariant.
So you can decompose $V=V_1\oplus\cdots\oplus V_k$, where the $V_i$ are $A/A^*$-eigenspaces, and are all $B$ and $B^*$-invariant. It follows that the restrictions $B_i\colon V_i\to V_i$ (more precisely, the 'compression') satisfies $B_i^*=B^*|_{V_i}$, so they are also normal.
By the spectral theorem, each $V_i$ has an orthonormal eigenbasis $\mathcal{B}_i$ for $B_i$. These will also be orthonormal eigenvectors for $A$, since $A$ acts as some multiple of the identity on each $V_i$. The union $\mathcal{C}=\mathcal{B}_i$ is an orthonormal eigenbasis both $A$ and $B$ on all of $V$. The matrix $U$ whose columns are the elements of this eigenbasis diagonalizes both $A$ and $B$.