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Aug 5, 2013 at 16:09 comment added Suvrit Yes, I later realized that I had forgotten the transposes which led me to write what I wrote.
Aug 5, 2013 at 15:37 vote accept Brian Lan
Aug 5, 2013 at 14:17 answer added Brian Lan timeline score: 1
Aug 5, 2013 at 13:18 comment added Brian Lan @suvrit: Let's take a simple example. Let $\mathbf{R} = \left[\begin{matrix}1 & 2\\ 0 & 1\end{matrix}\right]$. Then $\mathbf{Z} = \left[\begin{matrix}4 & 2\\ 2 & 0\end{matrix}\right]$ of which the eigenvalues are approximately $-0.8284$ and $4.8284$.
Aug 5, 2013 at 6:07 comment added Suvrit @Brian: I meant that in the case all matrices are real, then $Z = ND_R+D_RN+N^2$ has all zero eigenvalues (it is not symmetric however, so we are not talking about it being semidefinite, but having all zero eigenvalues is already nice; unfortunately, for complex matrices, this nice pattern breaks)
Aug 5, 2013 at 3:15 comment added Brian Lan Hi, suvrit. Thanks for your response. $\mathbf{Z}$ won't have zero eigenvalues. In fact, it won't be a positive semidefinite matrix, either.
Aug 5, 2013 at 2:42 comment added Brian Lan This is from a communication paper ieeexplore.ieee.org/xpl/…
Aug 4, 2013 at 4:42 comment added Suvrit Looks like a nice inequality and it seems that the triangular nature of $R$ is essential. If all matrices are real, then writing $R=D_R+N$, where $D_R$ is the diagonal and $N$ is the strict upper triangle we see that $RR^T=|D_R|^2+Z$, where matrix $Z$ has only zero eigenvalues; this should suffice to prove the statement; in the complex case, some more work is needed...
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Aug 3, 2013 at 15:30 history asked Brian Lan CC BY-SA 3.0