Timeline for A basic math identity often used in integrals [closed]
Current License: CC BY-SA 3.0
13 events
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| Jan 4, 2015 at 22:44 | history | closed |
DanielSank David Z |
Not suitable for this site | |
| Jan 4, 2015 at 22:38 | comment | added | DanielSank | This is pure math question with absolutely no physical context. | |
| Jan 4, 2015 at 22:36 | history | edited | JamalS | CC BY-SA 3.0 |
edited body
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| Sep 16, 2014 at 20:06 | comment | added | Valter Moretti | @Danu Sorry, I am presently too busy with lectures and other things to read and comment that question :c | |
| Sep 16, 2014 at 16:32 | vote | accept | LorentzNoether | ||
| Sep 16, 2014 at 8:39 | comment | added | MSalters | "Linear Algebra is linear". In the end, it's that simple. The multiplication is just a basis transform. Physical reality does not change by your arbitrary choice of a basis, so the (differential) equations describing reality should hold up in any basis. | |
| Sep 16, 2014 at 0:11 | answer | added | doetoe | timeline score: 5 | |
| Sep 15, 2014 at 22:45 | answer | added | ACuriousMind♦ | timeline score: 12 | |
| Sep 15, 2014 at 21:36 | comment | added | Valter Moretti | If $A$ is not a normal matrix it cannot be diagonalized that way. Instead $^T$ has to be replaced by $^{-1}$. However, in general $A$ is not diagonalizable and its determinant is not the product of its eigenvalues... | |
| Sep 15, 2014 at 21:27 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
deleted 11 characters in body; edited tags; edited title
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| Sep 15, 2014 at 21:21 | comment | added | Andrew McAddams | You may diagonalize $\hat{A}$: $\hat{A} = \hat{U}\hat{A}{'}\hat{U}^{T}$, where $det U = 1 , \hat{A}{'} = diag (A_{1}, ...)$. Then $det A = det U det A{'}det U^{T} = det A{'}$. | |
| Sep 15, 2014 at 21:21 | comment | added | higgsss | The Jacobian determinant! Actually it should be absolute value $|{\rm det} A|$ | |
| Sep 15, 2014 at 21:02 | history | asked | LorentzNoether | CC BY-SA 3.0 |