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:

$\rho(AB) \leq \rho(A)\rho(B)$ ?

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I've proved that if AB=BA the inequality is true. And better, if A and B are concurrently triangolable then the inequality is true. –  Ivan Feb 5 '13 at 15:41
Another proof: if $A$ and $B$ commute then $A$ and $B$ are co-diagonalizable and the inequality holds. –  Seirios Feb 5 '13 at 15:45
@Seirios You don't need $A$ and $B$ to be co-diagonalizable. That they commute is sufficient by the spectral radius formula. –  1015 Feb 5 '13 at 16:32

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

1. $A$ and $B$ are simultaneously triangularizable over $\mathbb{C}$. For instance, when $A$ and $B$ commute.
2. $A, B$ 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, $$\rho(AB)\le\|AB\|_2\le\|A\|_2\|B\|_2=\rho(A)\rho(B).$$
3. Both $A$ and $B$ are scalar multiples of row stochastic matrices, or both of them are scalar multiples of column stochastic matrices.
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Is this an "if and only if"? –  Ivan Feb 5 '13 at 23:44
@Ivan I don't think so. Those are a few circumstances that the inequality holds. There may be other possibilities. –  user1551 Feb 6 '13 at 4:45

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 $$\rho(C)=\lim_{n\rightarrow +\infty}\|C^n\|^{1/n}.$$

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We knew all this. –  Ivan Feb 5 '13 at 17:00
Well, apparently not everyone given certain answers and other claims in the comments. Plus I think the counterexample I gave is pretty straightforward. If you don't like it, don't upvote. –  1015 Feb 5 '13 at 17:03
You're right, excuse me. –  Ivan Feb 5 '13 at 17:08
No problem. Have a good day. –  1015 Feb 5 '13 at 17:12
Well, now the other answers and comments have disappeared, so the above comment looks rather foolish. –  1015 Feb 5 '13 at 19:14

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}

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