Let $A \in \mathbb{R}^n$ be a positive definite matrix. Then, it is well known that
\begin{align}
\text{Tr}(A^{-1}) \ge n^2 \text{Tr}(A)^{-1}
\end{align}
The proof follows by using the fact that trace is and a sum of eigenvalues and using AMGM inequality.
My question: Does this inequality hold with equality iff and only if $A$ is a diagonal matrix?
I know also that this inequality holds with equality iff eigenvalues of $A$ are identical. But not sure of this implies that $A$ is a diagonal matrix.
Best Answer
All positive-definite matrices are unitary diagonalizable $A=UDU^T$ and $$ \operatorname{Tr}A=\operatorname{Tr}UDU^T=\operatorname{Tr}DU^TU=\operatorname{Tr}D. $$ Similarly, $\operatorname{Tr}A^{-1}=\operatorname{Tr}D^{-1}$. Thus, it makes no restriction to assume that $A$ is diagonal. Of course, it is not going to be equality for a diagonal matrix in general (otherwise, it would be equality for all positive-definite matrices). It is equality iff all eigenvalues are equal, that is, iff $A=cI$, a scalar multiple of the identity matrix.
P.S. Just for completeness: the proof of the inequality follows immediately from Cauchy-Schwarz $$ \left(\sum\sqrt{\lambda_i}\cdot\frac{1}{\sqrt{\lambda_i}}\right)^2\le \sum\lambda_i\cdot\sum\frac{1}{\lambda_i} $$ with equality iff $(\sqrt{\lambda_1},\ldots,\sqrt{\lambda_n})$ and $(\frac{1}{\sqrt{\lambda_1}},\ldots,\frac{1}{\sqrt{\lambda_n}})$ are parallel, i.e. all $\lambda_i$ are equal.