Timeline for Natural explanation for a matrix identity
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 11, 2017 at 16:15 | comment | added | Fan Zheng | @FedorPetrov Edited for clarity. | |
| Oct 11, 2017 at 14:49 | history | edited | Fan Zheng | CC BY-SA 3.0 |
added 21 characters in body
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| Oct 11, 2017 at 14:35 | comment | added | Fedor Petrov | ah, the statement you ask about is in the next paragraph! This confused me. | |
| Oct 11, 2017 at 14:32 | history | edited | Fan Zheng |
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| Oct 11, 2017 at 14:27 | vote | accept | Fan Zheng | ||
| Oct 11, 2017 at 14:24 | comment | added | Fan Zheng | @FedorPetrov You're absolutely right in showing that $AS+SA$ is antisymmetric, so it can be used to produce a vector in the same way as $A$. | |
| Oct 11, 2017 at 12:34 | comment | added | Qfwfq | (I have corrected the typo: should read $-Sv$ instead of $-Sw$) | |
| Oct 11, 2017 at 12:33 | history | edited | Qfwfq | CC BY-SA 3.0 |
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| Oct 11, 2017 at 11:56 | answer | added | Neil Strickland | timeline score: 10 | |
| Oct 11, 2017 at 11:15 | comment | added | Somos | What if $A$ is the zero matrix? Then how could $AS+SA=0$ produce the vector $-Sw$? | |
| Oct 11, 2017 at 7:02 | comment | added | Fedor Petrov | Hm, $(AS+SA)^t=S^tA^t+A^tS^t=-SA-AS$, where I am wrong? | |
| Oct 11, 2017 at 5:16 | history | asked | Fan Zheng | CC BY-SA 3.0 |