Timeline for Determinant inequality of square-product sum of diagonal matrix and upper-triangular matrix
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
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... | |
Aug 3, 2013 at 15:48 | review | First posts | |||
Aug 3, 2013 at 15:55 | |||||
Aug 3, 2013 at 15:30 | history | asked | Brian Lan | CC BY-SA 3.0 |