Prove the inverse of a nonnegative matrix is nonnegative

Question

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?

Answer 1

Non-negative definite means that all its eigenvalues are greater than or equal to zero. Positive definite means that all its eigenvalues are greater than zero. Note that all positive definite matrices are also non-negative definite!

Think about what it means for a matrix to be invertible (in particular, how does its eigenvalues have to be) and you'll figure out why it has to be positive definite.

Answer 2

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.