# Matrix trace inequality proof

I am stuck on this problem. I have no idea how to proceed, so any help would be welcome.

Let $$A$$ be symmetric positive definite matrix $$n\times n$$ prove that : $$\operatorname{trace}(A) * \operatorname{trace}(A^{-1}) \ge n^2$$

• Hint: What is the relationship between trace and eigenvalues? What can you say about eigenvalues of a positive definite matrix? – Christian Sykes Sep 23 '18 at 18:45
• They are positive ? – AdnanM91 Sep 23 '18 at 18:49
• That answers the second question. What about the first? – Christian Sykes Sep 23 '18 at 18:50
• Sum of eigenvalues is same as trace – AdnanM91 Sep 23 '18 at 18:51
• Exactly. Finally, what are the eigenvalues of the inverse? You can put those three facts together to get what you need. – Christian Sykes Sep 23 '18 at 18:53

Hint: Let $$D$$ be a $$n\times n$$ diagonal matrix with eigenvalues $$0<\lambda_1 \leq \lambda_2 \leq \cdots \leq \lambda_n$$. Then we see that \begin{align} \operatorname{tr}(D)\operatorname{tr}(D^{-1}) =&\ \left(\lambda_1+\cdots +\lambda_n \right)\left(\frac{1}{\lambda_1}+\ldots+\frac{1}{\lambda_n} \right)\geq\ n^2 \end{align} where the last inequality follows from the AM-HM inequality.