Timeline for On a trace condition for positive definite $2\times 2$ block matrices
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2016 at 19:21 | vote | accept | Ludwig | ||
| Sep 18, 2016 at 18:01 | comment | added | Ruy | To get the strict inequality it is enough to apply the above solution to $X-\epsilon I$, where $\epsilon$ is small enough so as to make $X-\epsilon I$ positive. | |
| Sep 18, 2016 at 17:22 | comment | added | Suvrit | Alternatively, we know that $X > 0$ iff there exists a contractive matrix $K$ (i.e., $\|K\| < 1$) such that $C=A^{1/2}KB^{1/2}$. Thus, we see that \begin{equation*} \|C\|_F^2 = \mathrm{tr}(KAKB) < \text{rhs}, \end{equation*} because $K$ is a strict contraction. | |
| Sep 18, 2016 at 17:18 | comment | added | Ludwig | Yes, indeed, I'm wondering whether this still holds true with strict inequality. | |
| Sep 18, 2016 at 17:08 | comment | added | Suvrit | You are right, the OP asks for a strict inequality; I believe that it should be possible to make it strict without much trouble. | |
| Sep 18, 2016 at 16:46 | comment | added | T. Amdeberhan | But, the desired inequality is supposed to be sharp $<$ and not $\leq$. Am I right? | |
| Sep 18, 2016 at 16:29 | history | answered | Suvrit | CC BY-SA 3.0 |