Timeline for Is it true that $B \leq I \rightarrow B^2 \leq I$ (or higher powers) for positive semidefinite $B$, mimicking positive scalars? [closed]
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Apr 10, 2017 at 17:26 | history | closed |
R.P. Gro-Tsen Stefan Kohl♦ Jan-Christoph Schlage-Puchta Chris Godsil |
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| Apr 10, 2017 at 14:19 | answer | added | bmath | timeline score: 0 | |
| Apr 10, 2017 at 13:44 | review | Close votes | |||
| Apr 10, 2017 at 17:26 | |||||
| Apr 10, 2017 at 13:41 | comment | added | Fedor Petrov | Or simly note that all eigenvalues of $B$ are between 0 and 1, thus the same for $B^2$ | |
| Apr 10, 2017 at 13:23 | review | First posts | |||
| Apr 10, 2017 at 14:01 | |||||
| Apr 10, 2017 at 13:22 | comment | added | David Handelman | This looks like homework. But since $B$ has a positive square root, we can pre- and post-multiply by that square root, and obtain $B^2 \leq B$, hence $B^2 \leq I$. | |
| Apr 10, 2017 at 13:19 | history | asked | bmath | CC BY-SA 3.0 |