Let $B\in\mathcal{M}_{n\times n}(\mathbb{R})$ with positive entries. Prove that for $r>\rho(B)$ (spectral radius) then...

Question

the matrix $A=rI-B$ is nonsingular and $A^{-1}$ has nonnegative entries.

It's pretty easy to prove that $A$ is nonsingular since $$r\notin\text{spec}(B)\iff \det(rI-B)=\det(A)\neq 0.$$ But how do I go about showing that the entries of $A^{-1}$ are nonnegative? Thanks in advance!

Answer 1

We know that $B$ has positive entries by assumption. Also, note that $r>\rho(B)\geq0$. Now we wish to show that $A^{-1}$ has non-negative entries, where $A=rI-B$. Now,

\begin{align*} A^{-1} &= (rI-B)^{-1} \\ &= \frac{1}{r}\left(I-\frac{1}{r}B\right)^{-1} \\ &= \frac{1}{r} \sum_{k=0}^\infty \left(\frac{1}{r}B\right)^k \end{align*} We have that $\frac{1}{r^k}>0$ for all $k$, and $B$ has entries that are all positive so $B^k$ has all positive entries for all $k$ as well. Then, the sum of positive entries will of course remain positive. Hence, $A^{-1}$ has all non-negative entries.

Answer 2

Hint: Write $A^{-1}$ using an infinite series

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Second Hint: We have

$$ (I - M)^{-1} = \sum_{k=0}^\infty M^k $$ whenever the sum on the right converges (where we define $M^0 = I$ by convention). If $M$ has positive entries, then so does $M^k$ for $k \geq 1$.