TITLE: Spectral radius inequality
QUESTION [11 upvotes]: Suppose $A,B \in M(n \times n, \mathbb{C})$ or $ A,B \in M(n \times n, \mathbb{R}) $. Under wich hypothesis can I state that:
$\rho(AB) \leq \rho(A)\rho(B)$ ?

REPLY [10 votes]: (Edit: a number of conditions that I wrote down before are now merged into more general ones.) The inequality holds if:

$A$ and $B$ are simultaneously triangularizable over $\mathbb{C}$. For instance, when $A$ and $B$ commute.
$A, B$ are radial matrices. A complex matrix is called radial if its spectral radius coincides with its induced 2-norm (for example, all normal matrices are radial). When $A,B$ are radial,
$$\rho(AB)\le\|AB\|_2\le\|A\|_2\|B\|_2=\rho(A)\rho(B).$$
Both $A$ and $B$ are scalar multiples of row stochastic matrices, or both of them are scalar multiples of column stochastic matrices.