# Eigenvalues of n matrices

For $i \in \{1,...,n\}$, let $A_i=[a^i_{jk}]$ be a symmetric matrix. For $i \in \{1,...,n\}$, we assume that $a^i_{jj}=0$ for all $j$ and rank $(A_i)=2$. Then it has one positive and one negative eigenvalue.

In van Lint and Wilson "A course in combinatorics" page 79, they say that $A_1+\cdots+A_n$ can have at most $n$ positive (respectively negative) eigenvalues.

Could you tell me the proof?

• Have you tried with Weyl's inequalities? en.wikipedia.org/wiki/… ? That would be my first attempt. – Federico Poloni Jul 11 '18 at 6:44
• Is there any restriction on the size of the matrices, e.g. should they also be $n \times n$? – Vincent Jul 11 '18 at 8:03

## 1 Answer

If there are more than $n$ positive eigenvalues, the linear span of eigenvectors for $n+1$ of them is a space of dimension $n+1$ on which $A_1 + \ldots A_n$ is positive definite. There is then a nonzero vector $v$ in this space which, for each $i$, is orthogonal to the eigenvector of $A_i$ for its positive eigenvalue. But then $v^T A_i v \le 0$, resulting in $v^T (A_1 + \ldots + A_n) v \le 0$, contradiction.