$A\in M_n(\mathbb{R})$ symmetric matrix , $A-\lambda I$ is Positive-definite matrix- prove: $\det A\geq 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 $\lambda <a$. 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 $\lambda <a$. 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!
 A: $A-\lambda I$ is positive definite for every $\lambda<a$ means for all $\epsilon>0$, $A-(a-\epsilon) I$ 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.
A: Since $A$ is symmetric it is orthogonally diagonalizable and since orthogonal similarity preserves positive definitness, we may assume for all intents and purposes that $A$ is diagonal to begin with. We then know that $a_{ii}  - \lambda > 0$ for all $\lambda < a$ and $1 \le i \le n$, that is $a_{ii}  > \lambda$ for all $\lambda < a$ and for all $i$. It follows that $a_{ii} \ge a$ for all $i$, hence $\det A = \prod_{i=1}^{n}a_{ii} \ge a^n$.
A: I think you should not use $\lambda$ in $A-\lambda I$ because $\lambda$ usually denotes eigenvalues. So using $A-kI$ for example may be less confusing. 
Let $\mu_i(A)$ be the eigenvalues of $A$. Then the eigenvalues of $A-kI$ simply are $\mu_i(A)-k$. Since $A-kI$ is p.d., we have $\mu_i(A)>k, \forall k<a$. Hence $\mu_i(A)\ge a$ and consequently $\det A=\Pi\mu_i(A)\ge a^n$.
EDIT: I'm not sure whether we need use limitation to solve the problem as Sunni does.
