Timeline for Determinant of sum of positive definite matrices
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 26, 2019 at 17:06 | comment | added | Mark_Hoffman | @AndreasThom Thanks! | |
| Sep 26, 2019 at 6:03 | comment | added | Andreas Thom | I used $A(1+A^{-1/2}BA^{-1/2}) = A^{1/2}(A+B)A^{-1/2}$. For the second part, you would need to argue that the eigenvalues of $A^{-1}B$ are positive, which brings you back to my argument (I guess). | |
| Sep 25, 2019 at 8:30 | comment | added | Mark_Hoffman | @AndreasThom What did you do in the first step of your comment? I guess you expressed $A+B$ as a product, but I don't see why $A (A^{-1/2}B A^{-1/2})$ is equal to $B$ (if that is what it's used). Also, could you have started with $\det(A+B)=\det(A) \det(1+A^{-1}B)$ and follow from that? | |
| Apr 13, 2018 at 7:55 | comment | added | Andreas Thom | @F.Webber: It seems one has equality if and only if ${\rm rk}(A-B) \leq 1$. | |
| Apr 12, 2018 at 21:23 | comment | added | F.Webber | @AndreasThom How could one characterize the equality? | |
| Oct 10, 2017 at 18:44 | comment | added | Jose Brox | In fact, one can show a bit more with @AndreasThom idea: if $A$ is positive definite and $B$ is positive semidefinite and nonzero then $\det(A+B)>\det(A)+\det(B)$ (strict inequality), since $A^{-1/2}BA^{-1/2}$ is then nonzero positive semidefinite, hence with nonnegative eigenvalues and not all 0, and thus $\prod_i(1+\mu_i)\geq 1+\sum_i\mu_i+\prod_i\mu_i>1+\prod_i\mu_i$. | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| May 22, 2015 at 20:10 | comment | added | Jess Riedel | I'm going to disagree with @Suvrit and Andreas and say that the Minkowski theorem should be discussed. The reason is that it is a tighter bound that respects the dimensionality, and in particular is saturated when $A=B$ for any $n$. The OP's question is just an awkwardly weakened version of Minkowski. | |
| Jun 28, 2014 at 20:02 | comment | added | Andreas Thom | This is really misleading since it makes the question look like complicated, while is almost obvious that $\det(A+B) = \det(A) \det(1+A^{-1/2}BA^{-1/2}) \geq \det(A) (1+\det(A^{-1/2}BA^{-1/2})) = \det(A) + \det(B)$. In the second step, it is just used that $\prod_i (1+ \mu_i) \geq 1 + \prod_i \mu_i$, where $\mu_i$ are the eigenvalues of $A^{-1/2}BA^{-1/2}$. | |
| May 19, 2011 at 17:44 | vote | accept | user15221 | ||
| May 19, 2011 at 17:05 | comment | added | Suvrit | looks nice, but isn't it somewhat overkill to invoke the Minkowski theorem here? | |
| May 19, 2011 at 13:45 | history | edited | Andrey Rekalo | CC BY-SA 3.0 |
Added a link to a previous MO thread
|
| May 19, 2011 at 13:12 | history | edited | Andrey Rekalo | CC BY-SA 3.0 |
added 485 characters in body
|
| May 19, 2011 at 12:54 | history | answered | Andrey Rekalo | CC BY-SA 3.0 |