Timeline for Prove that absolute value of eigenvalue is smaller than 1 [closed]
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2021 at 1:52 | history | closed |
Christian Remling Michael Renardy Max Alekseyev Mikael de la Salle user44191 |
Not suitable for this site | |
| Jun 23, 2021 at 15:17 | answer | added | Denis Serre | timeline score: 2 | |
| Jun 23, 2021 at 15:13 | comment | added | Christian Remling | As for the question itself, write $A=D+B$, with $D$ the (block) diagonal part. You want to show that $-D^{-1}B$ has ev's $\le 1$. If $Bv=-\lambda Dv$, then $0=v^*(\lambda D+B)v= (\lambda-1)v^*Dv+v^*Av \ge (\lambda -1)v^*Dv$ and $v^*Dv\ge 0$ as well. | |
| Jun 23, 2021 at 15:11 | comment | added | Christian Remling | This is not the right site for this kind of question. Please use math.stackexchange.com instead. | |
| Jun 23, 2021 at 14:51 | review | Close votes | |||
| Jun 24, 2021 at 1:52 | |||||
| Jun 23, 2021 at 14:42 | comment | added | anonymousguyfromtheworld | I showed this to my professor and asked if this is true, as I need it for my thesis. He said it does indeed hold but did not explain the proof. | |
| Jun 23, 2021 at 14:37 | comment | added | Anthony Quas | Why do you think it's true? | |
| Jun 23, 2021 at 14:25 | review | First posts | |||
| Jun 23, 2021 at 14:29 | |||||
| Jun 23, 2021 at 14:21 | history | asked | anonymousguyfromtheworld | CC BY-SA 4.0 |