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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?

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  • $\begingroup$ Do you mean negative definite matrix? $\endgroup$
    – little o
    Feb 25, 2019 at 18:27

2 Answers 2

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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – John.Doh
    Feb 26, 2019 at 13:32

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