Timeline for If $A>B>0$, can we always find a positive real number $\alpha$, $0<\alpha < 1$ such that $\alpha A \geq B $?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 8, 2015 at 1:55 | answer | added | wayne | timeline score: 3 | |
| Jun 7, 2015 at 8:24 | vote | accept | wayne | ||
| Jun 7, 2015 at 5:17 | answer | added | Alex Ravsky | timeline score: 5 | |
| Jun 7, 2015 at 3:45 | comment | added | wayne | Thanks for you suggestion, after following your suggestion, I arrange it as follows: $A-B>\epsilon I$ which is equivalent to $A-\epsilon I> B$, also $(I-\epsilon A^{-1}) A> B$, but this still can't show $\alpha A>B$ where $0< \alpha <1$. | |
| Jun 7, 2015 at 3:29 | comment | added | Włodzimierz Holsztyński | The real part the relevant expression related to $\ A - B\ $ is greater than certain $\ \epsilon > 0\ $ when evaluated on the unit sphere. If you change $\ A-B\ $ a little, i.e. if you change $\ A\ $ a little then it'll be fine. | |
| Jun 7, 2015 at 3:10 | comment | added | wayne | If you mean my proposition is always true? | |
| Jun 7, 2015 at 3:08 | comment | added | Włodzimierz Holsztyński | Naturally. Thank you. Can't you prove a little sharper result: $\ \alpha\cdot A\ >\ B\ $ ? | |
| Jun 7, 2015 at 3:07 | comment | added | wayne | $A>B$ means $A-B$ is positive-definite matrix. | |
| Jun 7, 2015 at 2:43 | review | First posts | |||
| Jun 7, 2015 at 4:27 | |||||
| Jun 7, 2015 at 2:42 | history | asked | wayne | CC BY-SA 3.0 |