Let $A$ be symmetric positive definite matrix $n\times n$. Prove that $$\operatorname{trace} (A) \cdot \operatorname{trace} \left( A^{-1} \right) \ge n^2$$
I am stuck on this problem. I have no idea how to proceed, so any help would be welcome.
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Rodrigo de Azevedo
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AdnanM91
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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
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They are positive ? – AdnanM91 Sep 23 '18 at 18:49
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That answers the second question. What about the first? – Christian Sykes Sep 23 '18 at 18:50
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Sum of eigenvalues is same as trace – AdnanM91 Sep 23 '18 at 18:51
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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
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https://math.stackexchange.com/questions/391128/trace-of-an-inverse-matrix i have found this, ok I am now up to something. thanks! – AdnanM91 Sep 23 '18 at 18:58
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Related – Rodrigo de Azevedo Feb 16 '25 at 10:23
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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+\dots +\lambda_n \right)\left(\frac{1}{\lambda_1}+\dots+\frac{1}{\lambda_n} \right)\geq\ n^2 \end{align} where the last inequality follows from the AM-HM inequality.
Rodrigo de Azevedo
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Jacky Chong
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