linear algebra - $Ain M_n(mathbb{R})$ symmetric matrix , $A-lambda I$ is Positive-definite matrix- prove: $det Ageq a^n $

Let $a>1$ and $A\in M_n(\mathbb{R})$ symmetric matrix such that $A-\lambda I$ is Positive-definite matrix (All eigenvalues $> 0$) for every $\lambda

First, I'm not sure what does it mean that $A-\lambda I$ is positive definite for every $\lambda 0$ and all eigenvalues are bigger than $0$ or it's not.

Then, If it's symmetric I can diagonalize it, I'm not sure what to do...

Thanks!

Answer

$A-\lambda I$ is positive definite for every $\lambda0$, $A-(a-\epsilon) I$ is positive definite. That means, each eigenvalue of $A$ is larger than $a-\epsilon$, thus their product $\det A\ge \prod (a-\epsilon)$ ... let $\epsilon$ goes to zero, you get what you want.