Trace matrix inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T20:08:50Z http://mathoverflow.net/feeds/question/106231 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106231/trace-matrix-inequality Trace matrix inequality Josh 2012-09-03T10:41:45Z 2012-09-03T11:30:56Z <p>Hello all,</p> <p>I come across the following problem.</p> <p>Is it true that for a <strong>positive definite</strong> matrix $X^{n\times n}$, the following holds</p> <p>$\text{trace}(X^{-1})\geq\text{trace}([\text{diag}(X)]^{-1})$,</p> <p>where $\text{trace}(\;\cdot\;)$ and $\text{diag}(\;\cdot\;)$ are the matrix trace and diagonal operators, respectively.</p> <p>Remark 1: Note that in the above, there is equality when $X$ is diagonal.</p> <p>Remark 2: By using the SVD of the matrix $X$, or by using the Hadamard inequality, the above inequality is equivalent to</p> <p>$\sum\limits_{i=1}^n\frac{1}{\lambda_i}-\sum\limits_{i=1}^n\frac{1}{x_{ii}}\geq0$,</p> <p>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. </p> http://mathoverflow.net/questions/106231/trace-matrix-inequality/106233#106233 Answer by Mateusz Wasilewski for Trace matrix inequality Mateusz Wasilewski 2012-09-03T11:06:53Z 2012-09-03T11:06:53Z <p>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 <a href="http://en.wikipedia.org/wiki/Karamata%27s_inequality" rel="nofollow">Karamata's inequality</a> 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.</p>