# Prove this class of matrices has only one positive eigenvalue

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.

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

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

• Wow, that is not what I expected. I have investigated many of these matrices numerically, but have not found this counterexample. How did you construct/find this so quickly? – Wapiti Sep 27 '18 at 22:33
• I got it on the first try (for $n=5$ and entries in $[[1..10]]$). If it had not worked I would have increased $n$. The larger the number $n$, the more likely it is to find eigenvalues of different signums – loup blanc Sep 27 '18 at 22:37
• Is that Matlab? I wrote my own algorithm to explore but it explores all possible matrices, and so I only ran it up to 4,10 or so. – Wapiti Sep 27 '18 at 22:48

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.

• I knew most but not all of that. Interesting about the bounds! – Wapiti Sep 27 '18 at 23:06