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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...


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Hint: How do the eigenvalues of $A-\lambda I$ relate to those of $A$? So, what is the relationship between $a$ and the eigenvalues of $A$? – Srivatsan Aug 7 '11 at 12:47
You seem to be confused in the second paragraph. You are given that $A-\lambda I$ (and not the matrix $A$) is positive definite for all $\lambda < a$, which makes perfect sense. – Srivatsan Aug 7 '11 at 12:50
Thank you for the comment. I assume that if $\alpha$ is an eigenvalue of A so $\alpha - \lambda$ is an eigenvalue of $A-\lambda I$. You right regarding the confusion, I'll fix it. But, what does it mean to be positive define for onlt $\lambda < a$? – user6163 Aug 7 '11 at 13:18
up vote 2 down vote accepted

$A-\lambda I$ 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.

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Since eigenvalues are fixed, you could also simply say they are all${}\geq a$ and then multiply them. – Marc van Leeuwen May 26 '13 at 6:03

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 $\det A = \prod_{i=1}^{n}a_{ii} \ge a^n$.

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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.

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