Bounds on eigenvalues of symmetric $A$ where $I + A$ is positive-definite

Question

Let $A$ be a symmetric $n \times n$ matrix such that its diagonal entries are zero, and such that $I + A$ is positive-definite.

I'd like to show that the eigenvalues of $A$ are bounded from below by $-1$ and from above by $n - 1$.

The lower bound I can get from positive-definiteness of $I + A$ and the fact that its eigenvalues are sums of eigenvalues of $I$ and $A$.

But, I don't know where to start for the upper bound. Any hints?

Answer 1

Let $\lambda_1,\ldots,\lambda_n$ be the eigenvalues of $A$. Then the $1+\lambda_i$ are the eigenvalues of $I+A$, so are positive as $I+A$ is positive definite.

Also $\lambda_1+\cdots+\lambda_n=0$ as this is the trace of $A$. So $\lambda_1=-\lambda_2-\cdots-\lambda_n<1+\cdots+1=n-1$ etc.