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.

  • $\begingroup$ 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 $(I-B)^{-1} \neq I+B+B^{2}+\ldots$. $\endgroup$ – Brian Borchers Jan 29 at 16:29
  • $\begingroup$ @BrianBorchers I agree. $B^n$ can not converge to $0$ $\endgroup$ – Tony B Jan 29 at 18:32
  • $\begingroup$ So what happens if $B$ is a really big nonnegative matrix? $\endgroup$ – 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 ?

  • $\begingroup$ The inverse should be diag(2,-1). But I got your point. Your example works! From the economics point of view (input-output model), one industry is profitable, and the other is not. They don't interact with each other. $\endgroup$ – Tony B Jan 31 at 22:20
  • $\begingroup$ Yes, exactly. On the other hand, I go to correct the miscalculation. $\endgroup$ – loup blanc Jan 31 at 22:23

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