Timeline for $C\lVert\sum_i a_{ii}\rVert \ge \lVert(a_{ij})\rVert$ for matrices with entries in a $C^*$-algebra
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13, 2021 at 13:24 | vote | accept | Andromeda | ||
| Oct 31, 2021 at 20:18 | comment | added | Jamie Gabe | Ah, yes, this exercise in Paulsen's book is the correct generalisation of the n=2 case I mentioned above (the estimate I got applied the n=2 case over and over again to get a (quite bad) upper bound). This is much better! | |
| Oct 31, 2021 at 19:45 | comment | added | Andromeda | @JamieGabe Thanks for your comment. Please check out my answer. We can replace the factor $2^{n-1}$ by $n$. | |
| Oct 31, 2021 at 17:37 | comment | added | Jamie Gabe | Ah, I misread the the problem! Whoops! For the implication you want, you can use that $(a_{i,j})_{i,j=1}^n \leq 2^{n-1} \mathrm{diag}(a_1,\dots, a_n)$. The $2^{n-1}$ is definitely not optimal, but it works. To show the inequality, use the $n=2$ case over and over again (which is quite easy to prove) by considering matrices which "look like" $2\times 2$-matrices embedded in $n\times n$-matrices with a $n^2-4$ entries being zero. | |
| Oct 31, 2021 at 16:15 | comment | added | Andromeda | @JamieGabe This does not help. From your hint, we get $\|\sum_i a_{ii}\| \le n\|(a_{i,j})\|$ and we need the inequality in the other direction. | |
| Oct 31, 2021 at 14:37 | history | edited | LSpice | CC BY-SA 4.0 |
`\lVert\rVert`
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| Oct 31, 2021 at 12:15 | answer | added | Andromeda | timeline score: 3 | |
| Oct 31, 2021 at 11:45 | comment | added | Jamie Gabe | The map $M_n(A) \to A$ given by $(a_{i,j}) \mapsto \sum a_{i,i}$ is completely positive with norm $n$. | |
| Oct 31, 2021 at 10:30 | comment | added | Andromeda | @DiegoMartínez $C$ can depend on $n$. What did you have in mind? | |
| Oct 31, 2021 at 10:28 | comment | added | Diego Martinez | If you don't care if $C$ depends on $n$, then yes. If you do care, then not. | |
| Oct 31, 2021 at 9:50 | history | asked | Andromeda | CC BY-SA 4.0 |