# non-negative matrix satisfying two conditions

A real matrix $$B$$ is called non-negative if every entry is non-negative. We will denote this by $$B\ge 0$$.

I want to find a non-negative matrix $$B$$ satisfying the following two conditions:

(1) $$(I-B)^{-1}$$ exists but not non-negative. Here $$I$$ is the identity matrix.

(2) There is a non-zero and non-negative vector $$\vec{d}$$ such that $$(I-B)^{-1}\vec{d}\ge 0$$.

I tried all the $$2\times 2$$ matrices, but it did not work. I conjecture that such a $$B$$ does not exist, but don't know how to prove it.

• There are conditions under which $(I-B)^{-1}=I+B+B^2+\ldots$. In that case it's clear that $(I-B)^{-1}$ would be non-negative. Any counterexample would have be such that $(I-B)^{-1} \neq I+B+B^{2}+\ldots$. – Brian Borchers Jan 29 at 16:29
• @BrianBorchers I agree. $B^n$ can not converge to $0$ – Tony B Jan 29 at 18:32
• So what happens if $B$ is a really big nonnegative matrix? – Brian Borchers Jan 29 at 19:44

For example $$B=diag(1/2,2)$$; then $$(I-B)^{-1}=diag(2,-1)$$ and we can choose $$d=[1,0]^T$$.
Do you understand why such a $$B$$ works ?