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 : $$\operatorname{trace}(A) * \operatorname{trace}(A^{-1}) \ge n^2$$

  • $\begingroup$ Hint: What is the relationship between trace and eigenvalues? What can you say about eigenvalues of a positive definite matrix? $\endgroup$ Commented Sep 23, 2018 at 18:45
  • $\begingroup$ They are positive ? $\endgroup$
    – AdnanM91
    Commented Sep 23, 2018 at 18:49
  • $\begingroup$ That answers the second question. What about the first? $\endgroup$ Commented Sep 23, 2018 at 18:50
  • $\begingroup$ Sum of eigenvalues is same as trace $\endgroup$
    – AdnanM91
    Commented Sep 23, 2018 at 18:51
  • $\begingroup$ Exactly. Finally, what are the eigenvalues of the inverse? You can put those three facts together to get what you need. $\endgroup$ Commented Sep 23, 2018 at 18:53

1 Answer 1


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.


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