TITLE: Prove That $M^2+xM+yI$ and $M^2-xM+yI$ are non-singular
QUESTION [5 upvotes]: Let $M$ be an Invertible Hermitian matrix and let $x,y\in\Bbb R$ such that $x^2\lt 4y$,Then Prove That $M^2+xM+yI$ and $M^2-xM+yI$ are non-singular.
My Attempt:
$$(M^2+xM+yI)(M^2-xM+yI)=(M^2+yI)^2-(xM)^2$$
Now I Don't Know How to proceed further, I know that all the eigen values of Hermitian matrix are real. Help is needed.
Thank you.

REPLY [6 votes]: Complete the square wrt $M$ in both matrices
$$
M^2\pm xM+yI=\underbrace{\Bigl(M\pm\frac{1}{2}xI\Bigr)^2}_{\text{pos.semidef.}}+\underbrace{\frac{4y-x^2}{4}}_{\text{pos.def.}}I>0.
$$
The matrices are positive definite, hence, invertible.