# Necessary and sufficient condition for all the eigenvalues of a real matrix to be non-negative

According to wikipedia, a totally positive matrix (https://en.wikipedia.org/wiki/Totally_positive_matrix) always have non-negative eigenvalues. But, is it necessary for a matrix to be totally positive to have non-negative eigenvalues?

I'm looking for the necessary and sufficient condition for a real square matrix (not necessarily symmetric) to have non-negative eigenvalues.

• You may want to look up diagonal dominance and Gershgorin's circle theorems. – bcf Jun 3 '16 at 2:18
• @bcf : Thank you. Gershgorin's circle theorem looks like a good starting point. – Sachin Jun 3 '16 at 3:00

## 1 Answer

No it is not necessary. Just consider the matrix $$A= \begin{bmatrix} 2 & -1\\ -1 & 2 \end{bmatrix}$$ which is not totally positive but it is still has non-negative eigenvalues $\{1,3\}.$

• Is there any necessary condition then? – Sachin Jun 3 '16 at 5:35
• Yes, you can get many necessary conditions. e.g. for a square matrix $A,$ $det(A)\geq 0$ or $Tr(A) \geq 0$ are necessary conditions (although these are not sufficient) – A. Ray Jun 3 '16 at 5:44