Timeline for When does $\det(\frac{A+A^T}{2})=\det(A)$ for positive-definite $\frac{A+A^T}{2}$?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3 at 18:21 | comment | added | Dima Pasechnik | perhaps we are mixing up two types of similarity - od matrices, and of quadratic forms. | |
| Aug 3 at 18:02 | vote | accept | Aditya Bandekar | ||
| Aug 3 at 17:56 | comment | added | Branimir Ćaćić | This argument is correct. Indeed, since $B$ is assumed to the symmetric positive definite, it admits a unique symmetric positive definite square root $\sqrt{B}$ that you can construct via any orthogonal diagonalization of $B$; hence, take $g := \sqrt{B}$. @DimaPasechnik in general, $I = g^{-1} B g^{-t}$ need not be similar to $B$ precisely because $g$ need not be orthogonal. | |
| Aug 3 at 17:16 | comment | added | Aditya Bandekar | This looks correct. I'll think about it for a while and convince myself before accepting the answer. | |
| Aug 3 at 17:11 | comment | added | Joseph | Thanks for your comments. I have edited my answer to include more detail. | |
| Aug 3 at 17:10 | history | edited | Joseph | CC BY-SA 4.0 |
added 638 characters in body
|
| Aug 3 at 16:56 | comment | added | Dima Pasechnik | the best you can assume is that B is diagonal. | |
| Aug 3 at 16:55 | comment | added | Dima Pasechnik | B=I makes no sense - as any matrix conjugate to I equals I. | |
| Aug 3 at 16:23 | comment | added | Aditya Bandekar | He knows $B=gg^T=\frac{A+A^T}{2}$ is symmetric positive-definite, because he wrote $A$ in terms of its symmetric and antisymmetric parts. So $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$. The assumption given in the question is $B$ is positive-definite. I'm stuck on the line "Thus, by acting by this group, we can assume $B=1$". The rest of the proof I understand. Can someone make this more clear for me? | |
| Aug 3 at 16:19 | comment | added | Dima Pasechnik | Only symmetric p.d. matrices are of the form $gg^\top$. So your argument breaks down | |
| Aug 3 at 15:47 | history | edited | Joseph | CC BY-SA 4.0 |
edited body
|
| Aug 3 at 15:40 | history | answered | Joseph | CC BY-SA 4.0 |