Timeline for Bound the eigenvalue of product of matrices?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12, 2017 at 6:15 | comment | added | user31317 | I think I got it. | |
| Jun 12, 2017 at 5:41 | vote | accept | CommunityBot | ||
| Jun 12, 2017 at 4:03 | comment | added | user31317 | Can you explain in more detail why the geometric growth suffices? Can be proved using Jordan normal form? | |
| Jun 11, 2017 at 19:21 | comment | added | fedja | @user31317 I proved now that the norm growth is geometric, not just polynomial, which can be only if there is an eigenvalue bigger than one in absolute value. | |
| Jun 11, 2017 at 8:56 | comment | added | user31317 | Or since $H$ is invertible, then you claim $M$ has no eigenvalue with abs value 1? | |
| Jun 11, 2017 at 8:39 | comment | added | user31317 | Hello, so you proved that there exists a $x$ so that $\lim_{n \to \infty} |M^n x|$ diverges, but this is not sufficient for $M$ to have an eigenvalue with absolute value greater than $1$, but greater than or equal to one. | |
| Jun 10, 2017 at 3:29 | history | answered | fedja | CC BY-SA 3.0 |