Timeline for Commuting matrices of complex functions
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 6, 2020 at 19:26 | vote | accept | Guest | ||
| Apr 5, 2020 at 0:44 | history | edited | KhashF | CC BY-SA 4.0 |
Clarification added.
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| Apr 4, 2020 at 22:06 | comment | added | Guest | @Mark Yes. I hope so. | |
| Apr 4, 2020 at 21:40 | comment | added | user6976 | @KhashF: Perhaps you can add some clarification to your answer (about the transition from real to complex numbers). It could be that I am not the only one who got confused. | |
| Apr 4, 2020 at 17:59 | comment | added | user6976 | This makes sense. | |
| Apr 4, 2020 at 17:52 | comment | added | KhashF | @MarkSapir $A^\#(z)=(\overline{A(\bar{z})})^{\rm{T}}$ is holomorphic since you are conjugating twice (Schwarz Reflection); but $(A(z))^*=(\overline{A(z)})^{\rm{T}}$ is an anti-holomorphic function of $z$. | |
| Apr 4, 2020 at 17:35 | comment | added | user6976 | Sorry, my mistake is that I assumed that all functions involved are analytic. But the function $f(z)=\bar z$ is not analytic. Is it also a problem with the answer? | |
| Apr 4, 2020 at 14:57 | comment | added | KhashF | @MarkSapir I don't know what being "normal on the real line" means. Are you talking about a single matrix or a matrix-valued function? | |
| Apr 4, 2020 at 13:41 | comment | added | KhashF | @Guest I am saying if $z$ is real the matrices are normal and you have the desired equality. The equality persists throughout $\Bbb{C}$ due to Identity Theorem (en.wikipedia.org/wiki/Identity_theorem). | |
| Apr 4, 2020 at 5:53 | comment | added | Guest | @KhashF Do you mean that if $A(z) $ (with entire entries) is a normal matrix on the real line it will be normal on the whole complex plane? | |
| Apr 4, 2020 at 1:44 | comment | added | KhashF | @MarkSapir I think "entire" means analytic on $\Bbb{C}$: en.wikipedia.org/wiki/Entire_function | |
| Apr 4, 2020 at 1:34 | comment | added | KhashF | @MarkSapir Can you point out what's the mistake? I think $A(z)$ and $B(z)$ are normal once $z$ is real. I am only using the normality over the real line. | |
| Apr 3, 2020 at 23:57 | history | edited | KhashF | CC BY-SA 4.0 |
A mistake fixed; added 32 characters in body
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| Apr 3, 2020 at 23:54 | comment | added | KhashF | @Guest You are right, I edit my answer shortly. | |
| Apr 3, 2020 at 23:51 | comment | added | Guest | $A^{*}$ is not the same as $A^{\#} $. $A^{\#} (z) =A^{*} (\bar{ z}) $. | |
| Apr 3, 2020 at 23:46 | history | answered | KhashF | CC BY-SA 4.0 |