Timeline for Matrix function as gradient
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2023 at 8:34 | history | edited | Denis Serre | CC BY-SA 4.0 |
edited body
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| Jun 7, 2023 at 8:34 | comment | added | Denis Serre | @TitouanVayer Oh yes. Thanks. I edit my post. | |
| Jun 7, 2023 at 7:48 | comment | added | Titouan Vayer | sorry I meant for the condition $\langle \nabla_Y f(X), Z \rangle = \langle \nabla_Z f(X), X \rangle$ shouldn't it be $\langle \nabla_Y f(X), Z \rangle = \langle \nabla_Z f(X), Y \rangle$ instead ? | |
| Jun 7, 2023 at 7:27 | comment | added | Denis Serre | @TitouanVayer No, $X$ is an $X$. This means $\nabla_Yg(\det X)=g'(\det X)(\det X){\rm Tr}(X^{-1}Y)$. | |
| Jun 7, 2023 at 6:57 | comment | added | Titouan Vayer | Thank you ! in your first equation the last X is a Y right ? | |
| Jun 6, 2023 at 17:17 | comment | added | Denis Serre | @ChristianRemling this is exactly what I have written. Mind that $X$ is symmetric, so the transposition acts trivially. And teh factor is precisely the determinant itself. | |
| Jun 6, 2023 at 17:16 | history | edited | Denis Serre | CC BY-SA 4.0 |
added 59 characters in body
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| Jun 6, 2023 at 17:14 | comment | added | Denis Serre | @YCor Yes, this is the meaning. | |
| Jun 6, 2023 at 16:48 | history | answered | Denis Serre | CC BY-SA 4.0 |