# Spectral radius of the product of two matrices

Let us define the following matrix multiplication:

$$C=AB$$

where $B$ is a block diagonal matrix with $N$ blocks, $B_1$, $B_2$,..., $B_N$, each of dimensions $M \times M$. I know that $B_k = I_M - \mu R_k$ with $R_k$ equals to a hermitian matrix and $\mu$ equal to some positive constant. Moreover, I know that the the entries of the matrix $A$ are non-negative real numbers. I also know that the matrix $A$ is right stochastic, i.e., the sum of the elements in each row equals one. Can I say that the spectral radius of C is smaller than one for some values of $\mu$? If so, can I determine the range of values of $\mu$ under which the spectral radius of $C$ is smaller than one?

• Since $I$ is also Hermitian, can't you just say that $B_k$ is Hermitian? Given a Hermitian matrix $B$, you can compute $X = I-B$, which is also Hermitian, and call $X$ by the name $R_k$, so simply saying $B_k$ is Hermitian is the same as saying it has the form you specified. Feb 20 '16 at 15:00
• True. You are right $X=I-B$ is hermitian. Does this help to have an answer? Thanks! Feb 20 '16 at 15:09
• No idea. But it does simplify the question, which is sometimes a good start. And it makes user1551's counterexample no longer work in general, because the matrix $B$ produced by that answer is generally not Hermitian. Feb 20 '16 at 15:34

The answer is no in general. Suppose $A$ is stochastic but not doubly stochastic. Then its operator norm (i.e. the largest singular value) is strictly greater than $1$. Let $A=USV^T$ be its singular value decomposition. Let $0<\frac1{\|A\|_2}<\epsilon<1$ and define $B=\epsilon VU^T$. Then $B$ itself is a big $n\times n$ block matrix whose operator norm is equal to $\epsilon<1$. Yet the spectral radius of $AB=\epsilon USU^T$ is $\epsilon\|A\|_2$, which is greater than $1$.
By extending the above construction, we can construct counterexamples such that both $A$ and $B$ are block diagonal and they have identical partitions. Therefore, by continuity, there exist counterexample where the $A$ is entrywise nonzero but $B$ is genuinely/properly block-diagonal as well.
• In particular for $A = \begin{bmatrix} 0 & 1 \\ 0 & 1\end{bmatrix}, B = \begin{bmatrix} s & -s \\ s & s\end{bmatrix}$, where $s = c \sqrt{2}/2$, we get eigenvalues $c\sqrt{2}$ and $0$ for $AB$. In this case, the 2-norm of $B$ is $c$, so picking any value of $c$ like $0.8$ provides a numerical example. Feb 20 '16 at 13:40
No. Consider $$B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\\ A = \begin{bmatrix} -10 & 9 \\ 9 & -10 \end{bmatrix}$$ The determinant of $AB$ is $19$, so at least one of the two eigenvalues must be larger than $1$. (They're real because $AB$ is symmetric.)
If you also know that all entries in $A$ are positive, then I suspect that it might be true, but don't have a proof offhand.