# Strictly diagonally dominant hermitian matrices eigenvalues sign

Let $A\in \mathcal{M}_{n\times n}(\mathbb{C})$ be a strictly diagonally dominant hermitian matrix.

My main goal is to tell how many positive eingenvalues $A$ has in terms of its leading diagonal entries $a_{ii}$.

To do this it suffices to show that every Gershgorin disc contains at least one eigenvalue.

And to prove the above statement it suffices that any two Gershgorin discs do not intersect. But I'm not sure the last statement true, nor can I prove it.

Any help?

## 1 Answer

You do not need to show that the discs are disjoints. In fact, this won't hold for most diagonally dominant matrices, unlike the main result that you wish to prove.

What you need is a stronger form of the Gerschgorin disc thorem, which is due to O. Taussky-Todd and is today normally taught alongside the standard version:

Theorem If the union of $k$ Gerschgorin discs is disjoint from the union of the other $n-k$ discs then the former union contains exactly $k$ and the latter $n-k$ eigenvalues of $A$. (Wikipedia reference with proof).

• I'm aware of that result, it was presented to me as being part of Gershgorin's theorem. And as I write this to you, I realize that what I want to prove follows trivially from that. Thanks. Dec 30, 2012 at 15:41