Timeline for A generalized log inequality for positive definite trace-one matrices
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 12, 2017 at 16:59 | vote | accept | Ludwig | ||
| Feb 12, 2017 at 16:57 | comment | added | jjcale | I added the trace. From $x \ge x^2$ it follows $x \le 1$ . In our case $x = tr \sum_{i=1}^N\mu_i$ . | |
| Feb 12, 2017 at 16:51 | history | edited | jjcale | CC BY-SA 3.0 |
added 3 characters in body
|
| Feb 12, 2017 at 16:14 | comment | added | Ludwig | I think that in the last equality there should be a trace, namely $\left(\mathrm{tr}\sum_{i=1}^N\mu_i\right)^2$ instead of $\left(\sum_{i=1}^N\mu_i\right)^2$. Furthermore, I don't see why the latter expression should be greater or equal to 1 (in order to conclude the proof). Thanks again! | |
| Feb 12, 2017 at 13:36 | comment | added | jjcale | For positive definite matrix A it follows $log A \ge I - A^{-1}$ . | |
| Feb 12, 2017 at 13:08 | comment | added | Ludwig | I'm still missing something: How does the inequality $\log x \ge 1-1/x$ apply to the case of the trace of the matrix logarithm? | |
| Feb 12, 2017 at 12:54 | comment | added | jjcale | @Jacquard : I apply the inequality to (2) after moving the factor $(V_i^T X V_i)^{-1/2}$ to the left in order to make the operand symmetric. | |
| Feb 12, 2017 at 11:05 | comment | added | Ludwig | Thanks for the answer! Could you please elaborate a little more on the equivalent inequality $\mathrm{tr}\sum_{i=1}^N\mu_i\leq 1$? More precisely, how did you use the inequality $\log x \geq 1-1/x$ in order to derive the latter expression? | |
| Feb 12, 2017 at 10:26 | history | answered | jjcale | CC BY-SA 3.0 |