Perron root of convex combination of nonnegative matrices

Question

Given (irreducible) nonnegative matrices $A_i$ and a convex weighting $\hat{w}$ (i.e. a nonnegative set of reals ${w}$ s.t. $\sum_i w_i$) what can we deduce about the Perron root of $\sum_i w_i A_i$, in terms of the Perron roots of $A_i$? The classical theorems require $A_i$ to all be symmetric (Hermitian) or diagonal.

What can we say about bounds for this Perron root? Approximations?

Answer 1

Given that the dominant eigenspace of each component (nonnegative irreducible) matrix is known, the results of L. Yu. Kolotilina apply here. In particular, to restate their Theorem 5:

$$ \rho\left(\sum_i A_i\right) \geq \sum_i \alpha_i \rho(A_i) $$

in which the coefficients $\alpha_i$ are given by a product of ratios of the left and right eigenvectors for $A_i$ as can be seen on page 10. Within, they also discuss the case of equality in which all left and right eigenvectors of $A_i$, namely $u^{(i)}, v^{(i)}$ satisfy $u^{(i)} \cdot v^{(i)} = c \in \mathbb{R}^N$.

Answer 2

In the general case, not much, I am afraid. You have obviously $\rho(\sum_i A_i)\geq \max_i \rho(A_i)$ but that seems to be all. In particular, you don't have $\rho(\sum_i A_i)\leq \sum_i \rho(A_i)$ as you would if the matrices were symmetric.

For example (letting $\epsilon\approx 0$ be a small positive number) with $A_1=\pmatrix{\epsilon & 1\\ \epsilon& \epsilon}$ and $A_2=A_1^t$ you have $\rho(A_i)\approx 0$ but $\rho(A_1+A_2)\approx 1$ while for $A_1=\pmatrix{1 &\epsilon\\ \epsilon& \epsilon}$, $A_2=\pmatrix{\epsilon & \epsilon\\ \epsilon& 1}$, you have $\rho(A_i)\approx 1 \approx \rho(A_1+A_2)$.

EDIT: For the lower bound stated in the beginning of my post, it may be shown from the Collatz-Wielandt characterization of the Perron value.

In dim $d$, one sets for $x\in {\Bbb R}_+^d$, $x\neq 0$ and understanding that $\inf\{+\infty,a\}=a$: $$ \lambda_A(x) = \inf_i \frac{(Ax)_i}{x_i}$$ Then C-W tells us that $\rho(A)=\max\{ \lambda_A(x) : x\in {\Bbb R}_+^d, |x|=1\}$. For positive matrices $A$ and $B$ we clearly have $\lambda_{A+B}(x) \geq \max\{ \lambda_A(x), \lambda_B(x)\}$ and the claim follows.