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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.

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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.

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Prove that if a matrix $A$ is diagonalizable with distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ and if $p(t) = \sum_{i=0}^n a_i t^i$ is a polynomial (with $a_i \in \mathbb{F}$), then $p(A)$ is also diagonalizable with eigenvalues $p(\lambda_1), \ldots, p(\lambda_k)$ (some of them may repeat). Apply this to $M$ and $p(t) = t^2 + xt + y$ and check that the eigenvalues of $p(M)$ are all non-zero to guarantee that $p(M)$ is non-singular.

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  • $\begingroup$ All the Hermitian matrix are diagonalizable, So $M$ is diagonalizable, also it is non singular hence all the eigen values are non zero $\endgroup$ – Chiranjeev_Kumar Nov 7 '15 at 16:07
  • $\begingroup$ Is there any other approach to handle this problem? $\endgroup$ – Chiranjeev_Kumar Nov 7 '15 at 16:21
  • $\begingroup$ There are many approaches, but in my opinion they are more tricky and less general. You can let $A = M^2 + xM + yI$ and show by direct calculation using Cauchy-Schwartz inequality and the fact that $M$ is Hermitian that $\left<Av, v \right> > 0$ for $v \neq 0$. This shows that (the linear map corresponding to) $A$ is injective and thus invertible. $\endgroup$ – levap Nov 7 '15 at 16:31

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