# Prove the inverse of a nonnegative matrix is nonnegative

Defintion of a nonnegative matrix:

Symmetrical matrix $$A: n \times n$$ is non-negatively defined when $$A > 0$$ or $$A ≥ 0$$

We have to prove the following: If $$A$$ is defined as a nonnegative matrix, there exists $$A^{-1}$$ only while $$A>0$$.

I don't understand how to prove this property. Why am I told $$A$$ has been defined as nonnegative and yet for an inverse to exist it has to be positive?

• Do you mean negative definite matrix? Feb 25, 2019 at 18:27

because a symmetric, $$A$$, can be written as $$U^{*}DU$$. And we have known that there is no 0-entry in $$D$$, so we can get $$A^{-1}$$ = $$U^{*}D^{-1}U$$. Or you can just say that the $$dim(null(D)) = 0$$ $$\implies$$ $$dim(null(A))= 0$$ $$\implies$$ $$A$$ is invertible.
• I understand that for a matrix to be invertible the number $0$ must not be an eigenvalue of $A$. However I do not understand how this helps me prove that an inverse matrix exists. Feb 26, 2019 at 13:32