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 . I need to prove that $\det A\geq a^n $.
First, I'm not sure what does it mean that $A-\lambda I$ is positive definite for every $\lambda . It's whether $A>0$ 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!