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Let $A \in \mathbb{R}^n$ be a positive definite matrix. Then, it is well known that

$$ \mbox{Tr} \left( A^{-1} \right) \ge n^2 \, \mbox{Tr}(A)^{-1} $$

The proof follows by using the fact that trace is and a sum of eigenvalues and using AM-GM 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.

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  • $\begingroup$ I think the above relation is wrong in the sense that it should have been $n$ instead of $n^2$ on the right hand side. Remember that there is a square-root term inside the frobenius norm, so the square of the frobenius norm will just be the size of the identity Matrix and not the square of the size of the identity Matrix. $\endgroup$ Nov 20, 2022 at 6:27

2 Answers 2

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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.

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Edit : I missed the assumption that the matrix should be symmetric.


There is equality if and only if all the eigenvalues of $A$ are equal. This does not imply that $A$ is diagonal. For instance, take $$A = \begin{pmatrix} 1&1\\0&1\end{pmatrix} \, . $$ It is not diagonal but saturates your inequality.

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  • $\begingroup$ Can we say something about the matrix? Rather than it's eigen values. $\endgroup$
    – Boby
    Mar 24, 2017 at 15:07
  • $\begingroup$ Well, you can say that its eigenvalues are all equal, or any other equivalent statement. No more. $\endgroup$
    – Antoine
    Mar 24, 2017 at 15:08
  • $\begingroup$ Do you know if this class of matrices has a name? $\endgroup$
    – Boby
    Mar 24, 2017 at 15:09
  • $\begingroup$ For instance $\begin{pmatrix} 38&25\\-1&28\end{pmatrix}$ has two equal eigenvalues, and it is not diagonal, not triangular, etc. I'm not aware of any particular name. $\endgroup$
    – Antoine
    Mar 24, 2017 at 15:11
  • $\begingroup$ The matrix $A$ is assumed to be positive-definite, hence, symmetric. Your example is not a symmetric matrix. $\endgroup$
    – A.Γ.
    Mar 24, 2017 at 15:15

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