Prove this class of matrices has only one positive eigenvalue

Question

I am working with real symmetric non-negative matrices with integer elements and zero diagonal. They are particularly nice, and I am fairly sure that they all have exactly one positive eigenvalue.

Sometimes they can be singular, for example

$$A = \left[ \begin{matrix}0,1,1,1 \\ 1,0,1,1 \\ 1,1,0,4 \\ 1,1,4,0\end{matrix}\right]$$

But still, only one positive eigenvalue. Is there a way to show that these matrices have exactly one positive eigenvalue generally? I thought that proving them conditionally negative definite would help, as here, but I was wrong.

Answer 1

It's false; it suffices to randomly choose an instance of such a matrix.

$Matrix(5, 5, [[0, 2, 8, 4, 2], [2, 0, 5, 9, 7], [8, 5, 0, 3, 10], [4, 9, 3, 0, 10], [2, 7, 10, 10, 0]])$

Answer 2

HINT:

I can just guess which properties are common to your matrices. I assume:

They are symmetric, and so all eigenvalues are real.

They have zero main diagonal and positive non-diagonal elements. Therefore they are primitive, and all statements of the Perron-Frobenius theorem are valid for them. In particular, the eigenvalue with greatest absolute value is positive, and thus equal to the spectral radius.

This leading eigenvalue $r$ is greatest than the minimum of row sums, and is smaller than the maximum of row sums.

In the case of the matrix you wrote, $3\leq r \leq 6.$

Since the diagonal entries are all zeros, the row sums are also radii of Gerschgorin circles centered in $0.$

The sum of eigenvalues is $0,$ as it is the trace.