Timeline for Maximize function on rotation matrices [closed]
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2020 at 11:29 | history | closed |
user44191 Alex M. Henry.L Dima Pasechnik Nemo |
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| Apr 17, 2020 at 7:23 | answer | added | Federico Poloni | timeline score: 4 | |
| Apr 17, 2020 at 7:18 | history | edited | Benjamin Techer | CC BY-SA 4.0 |
added 36 characters in body
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| Apr 17, 2020 at 7:12 | history | edited | Benjamin Techer | CC BY-SA 4.0 |
added 36 characters in body
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| Apr 17, 2020 at 7:07 | comment | added | Benjamin Techer | Thanks @leomonsaingeon for you answer. I will try to make my question clearer. I fix $A$ (which is in my case the tensor of velocity gradient at point $(x,y,z)$ in the space), then I will look for a new frame, that consists of rotating the laboratory frame (to get a new frame $x*=Qx$), for which the new velocity gradient $B=QAQ^\top$ maximizes F. Practically, $F$ depends on $\phi$, $\theta$ and $\psi$ as variable and $A$ as a fixed matrix. | |
| Apr 17, 2020 at 6:28 | comment | added | leo monsaingeon | Welcome to MO. You question is unclear and needs more explanations: Do you fix $A$ and look at $F=F(Q)$ as a function of $Q$? Or on the contrary do you fix $Q$ and look at $F=F(A)$ as a function defined on the whole vector space of matrices? In the first case your question makes no sense, because the set of rotation matrices is not convex. In the second case the answer is no, there is no concavity (take $Q=Id$ and look matrices with zero coefficients except $A_{12}$, in which case $F(A)=\frac 12 |A_{12}|^2$ is clearly not concave). I vote as off-topics, not research level. | |
| Apr 17, 2020 at 0:20 | review | Close votes | |||
| Apr 21, 2020 at 11:37 | |||||
| Apr 16, 2020 at 23:42 | review | First posts | |||
| Apr 17, 2020 at 0:06 | |||||
| Apr 16, 2020 at 23:35 | history | asked | Benjamin Techer | CC BY-SA 4.0 |