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It's easy to proof by counterexample that non-negative matrices can have negative eigenvalues.

For example, the following matrix has -1 as an eigenvalue:

$$ A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} $$

However, which are the properties of those matrices, is there a generalization of them?

Thanks!

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For a real-valued and symmetric matrix $A$, then $A$ has negative eigenvalues if and only if it is not positive semi-definite. To check whether a matrix is positive-semi-definite you can use Sylvester's criterion which is very easy to check.

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  • $\begingroup$ You right it only holds for symmetric matrices, I'll edit my answer. Thanks! $\endgroup$ – Snufsan Jul 28 '16 at 8:44
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If your matrix is invertible and positive, then either it already has at least one negative eigenvalue, or you can get a matrix with a negative eigenvalue by exchanging two rows.

Proof:

If all eigenvalues are positive, then the determinant is positive. Exchanging two rows changes the sign of the determinant. Since the determinant is the product of the eigenvalues, a matrix with a negative determinant has at least one negative eigenvalue.

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For $(2,2)$ matrices with positive entries the following are equivalent

  1. the determinant non positive
  2. the matrix has on negative eigenvalue.

Note that the sum of the two eigenvalues is positive whereas their product is negative : such a matrix must be diagonalizable, and the two eigenvalues must have opposite sign.

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