Spectral radius inequality 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:
$\rho(AB) \leq \rho(A)\rho(B)$ ?
 A: It is not true in general that $\rho(AB)\leq \rho(A)\rho(B)$.
Consider:
$$
A=\left( \matrix{1&0\\ 1& 1}\right)\quad B=\left( \matrix{1&1\\ 0& 1}\right)
$$
Then $\rho(A)=\rho(B)=1$.
But
$$
AB=\left( \matrix{1&1\\ 1& 2}\right)
$$
has $\rho(AB)=(3+\sqrt{5})/2$.
If $A$ and $B$ commute, we have
$$
\|(AB)^n\|=\|A^nB^n\|\leq \|A^n\|\|B^n\|
$$
hence
$$
\|(AB)^n\|^{1/n}\leq \|A^n\|^{1/n}\|B^n\|^{1/n}.
$$
Letting $n$ tend to $+\infty$, we find the desired inequality thanks to the Spectral Radius Formula (or Gelfand's formula): http://en.wikipedia.org/wiki/Spectral_radius
$$\rho(C)=\lim_{n\rightarrow +\infty}\|C^n\|^{1/n}.$$
A: (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.

A: I've found an interesting counterexample for inequality
A = \begin{bmatrix} 5 & 2 & 1\\ 4 & 0 & 0\\ 3 & 0 & 1\end{bmatrix}
B = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 1\\ 0 & 0 & 1\end{bmatrix}
