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Hello all,

I come across the following problem.

Is it true that for a positive definite matrix $X^{n\times n}$, the following holds

$\text{trace}(X^{-1})\geq\text{trace}([\text{diag}(X)]^{-1})$,

where $\text{trace}(\;\cdot\;)$ and $\text{diag}(\;\cdot\;)$ are the matrix trace and diagonal operators, respectively.

Remark 1: Note that in the above, there is equality when $X$ is diagonal.

Remark 2: By using the SVD of the matrix $X$, or by using the Hadamard inequality, the above inequality is equivalent to

$\sum\limits_{i=1}^n\frac{1}{\lambda_i}-\sum\limits_{i=1}^n\frac{1}{x_{ii}}\geq0$,

where $\lambda_i$ and $x_{ii}$ are the $i$th eigenvalue of the matrix $X$, and the $i$th diagonal element of the matrix $X$, respectively.

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1 Answer 1

up vote 13 down vote accepted

It is quite well known that the diagonal of an Hermitian matrix is majorized by sequence of its eiqenvalues (this is known as Schur's theorem). Since all of these numbers are positive, we can use Karamata's inequality for function $x \mapsto \frac{1}{x}$ to conclude that inequality $\sum_{i=1}^{n} \frac{1}{\lambda_{i}} \geqslant \sum_{i=1}^{n} \frac{1}{x_{ii}}$ is true.

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Great! Thank you! –  Josh Sep 3 '12 at 11:41
    
You're welcome. –  Mateusz Wasilewski Sep 3 '12 at 12:35

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