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!
$\endgroup$3 Answers
$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$For $(2,2)$ matrices with positive entries the following are equivalent
- the determinant non positive
- 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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