Timeline for Showing a matrix is negative definite [formerly Showing a sum is always positive]
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 29, 2010 at 12:57 | comment | added | fedja | Subtracting $A^2+B^2$ from both sides, we get $2AB\le \frac{d-1}{d}A^2+\frac{d}{d-1}B^2$, which is pura AM_GM | |
| Mar 29, 2010 at 11:27 | comment | added | bandini | Sorry, one more question. Could you provide justification for the inequality $(A+B)^2 \leq \frac{2d-1}{d}A^2 + \frac{2d-1}{d-1}$? | |
| Mar 29, 2010 at 2:03 | vote | accept | bandini | ||
| Mar 29, 2010 at 0:49 | history | answered | fedja | CC BY-SA 2.5 |