Timeline for Closed form solution for $XAX^{T}=B$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2022 at 7:33 | comment | added | Federico Poloni | In any case, the non-symmetric case is much more involved as far as I know; a good starting point is: Canonical forms for congruence of matrices: a tribute to H. W. Turnbull and A. C. Aitken. F. De Terán. SeMA Journal, 73 (2016) 7-16. | |
| Sep 16, 2022 at 7:28 | comment | added | Federico Poloni | @wikiwert AFAIK the standard definition of "positive definite" includes "symmetric"; if you mean "Hurwitz anti-stable" (all eigenvalues in the right half-plane) please specify it. | |
| Sep 15, 2022 at 14:13 | comment | added | wikiwert | The question asks about positive definite matrices, not symmetric matrices. However, if $A^{1/2}$ is a symmetric square root, $A^T = (A^{1/2}A^{1/2})^T=A$, so $A$ is also symmetric. The given solution only works if $A$ and $B$ are symmetric. Unfortunately Federico Poloni's answer doesn't answer the question in general as I thought. If the question intended to ask about symmetric matrices, maybe it should be edited. | |
| May 29, 2021 at 7:29 | comment | added | user_na | And with that, $X=MQL^{-1}$, right? | |
| May 29, 2021 at 7:15 | comment | added | Federico Poloni | @user_na If $B=MM^T$ and $A=LL^T$, then following the same steps one gets that $M^{-1}XL=Q$ satisfies $QQ^T=I$. | |
| May 29, 2021 at 6:46 | comment | added | user_na | @federico-poloni thank you, this was very helpful, could you elaborate on how the Cholesky factor approch would looke like? | |
| Jul 20, 2020 at 16:01 | vote | accept | Fabio | ||
| Jul 20, 2020 at 15:58 | history | answered | Federico Poloni | CC BY-SA 4.0 |