Eigenvalues of the product of two diagonalizable commuting matrices. Let $A, B$ be two n by n matrices. Suppose that $A, B$ are diagonalizable and $AB=BA$. Do we have the following: eigenvalues of $AB$ are the products of eigenvalues of $A, B$?
Edit: Do we have the following: eigenvalues of $AB$ are products of eigenvalues of $A, B$?
 A: Since $A$ and $B$ commute, they can be diagonalized simultaneously, i.e. there is an invertible matrix $S$ such that
$$ A = S^{-1} \mathrm{diag}(\lambda_1, \dots, \lambda_n) S, \quad B = S^{-1} \mathrm{diag}(\mu_1, \dots, \mu_n) S.$$
where $\mathrm{diag}$ is a diagonal matrix with the specified diagonal entries (the eigenvalues of $A$ resp. $B$). Then
$$ AB = S^{-1} \mathrm{diag}(\lambda_1, \dots, \lambda_n) SS^{-1} \mathrm{diag}(\mu_1, \dots, \mu_n) S = S^{-1}\mathrm{diag}(\lambda_1, \dots, \lambda_n)\mathrm{diag}(\mu_1, \dots, \mu_n) S \\ = S^{-1}\mathrm{diag}(\lambda_1 \mu_1, \dots, \lambda_n \mu_n) S $$ 
and the eigenvalues are indeed, as you suspected, the product of the eigenvalues. Observe that the order in which the eigenvalues are multiplied depend on the matrix $S$, i.e. is not arbitrary.
A: $A,B$ do not need to be diagonalizable.
Proposition. Let $A,B\in M_n(\mathbb{C})$ and $spectrum(A)=(\lambda_i)_i$. 
If $AB=BA$, then, there is an ordering of $spectrum(B)$: $(\mu_i)_i$, s.t. $spectrum(A+B)=(\lambda_i+\mu_i)_i$ and $spectrum(AB)=(\lambda_i\mu_i)_i$.
Proof. $A,B$ are simultaneously triangularizable.
