It is well known that for an endomorphism $\alpha: V \to V$ with $V$ a finite-dimensional inner-product space over $\mathbb{C}$, $\alpha$ has purely real eigenvalues. But if we want to bound these eigenvalues in some range $[a,b]$, the only result I could find on this is that all eigenvalues $\lambda_{i}\in [a,b]$ $\iff$ $\alpha-c I$ is positive definite for all $c<a$ and negative definite for all $c>b$. However, I can't seem to prove this (and my source of this contains no proof either).
I was wondering if anyone had seen this result before, and has any idea how to prove it.
Hints: A self-adoint endomorphism is positive if all its eigenvalues are positive. The eigenvalues of $\alpha$ are $\lambda_1,\dots,\lambda_k$ iff the eigenvalues of $\alpha - cI$ are $\lambda_1 - c, \dots, \lambda_k - c$. This implies that the eigenvalus of $\alpha$ lie in $[a,b]$ iff the eigenvalues of $\alpha - cI$ lie in $[a - c, b - c]$. What does this say on the eigenvalues of $\alpha - cI$ if $c < a$ or $c > b$?