Timeline for The right expansion of a square root matrix
Current License: CC BY-SA 3.0
12 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Mar 9, 2015 at 17:34 | comment | added | user68601 | Thank you again for the reply. I have tried to ask clarifications about the correctness of the answer provided in the second link. Unfortunately, the way I managed things were not suitable with the site's policies and so I was forced to ask a new question (the one asked here). In any case I would like to hear Carlo's derivation of that formula in order to be sure that we are not missing anything at all here and, if needed, to adjust the reply appeared in the the second link. By the way Igor: essentially you are telling me that the answer in the second link is wrong, isn't it? | |
| Mar 9, 2015 at 16:32 | comment | added | Igor Khavkine | OK, I just hope that you have realized by now that the formula $D=A^{-1/2}B/2$ is NOT correct, unless $A$ and $B$ are simultaneously diagonalizable. You may have to face the fact that the solution for $D$ may have no formula that you would consider "nice". If there is no single "nice" way to write the answer, there will always be an infinitude of possibly ugly ways to write the answer, and you'll just have to choose one of them depending on the context of your problem. | |
| Mar 9, 2015 at 13:14 | comment | added | user68601 | @Igor: you are right about missing information in the present post and I apologize for the inconvenience. The problem I need to solve is (hopefully) well stated in the first link and is the following: I need an expansion of $(A+n^{-1/2}B)^{1/2}$ in terms of solely $A$ and $B$. The Lyapunov equation just came out as a device to solve the problem. As I am not a mathematician and because Carlo's reply (the one in the second link) meet my requirements, I was trying to figure out if there exists any relation between the solution to $B=A^{1/2}D+DA^{1/2}$ and Carlo's answer. | |
| Mar 9, 2015 at 11:41 | comment | added | Igor Khavkine | @user68601 (if that's your real name :-), I don't see any information in the question about what would or would not be feasible in your calculation. I think the following question has been answered: Does the equation $B=\sqrt{A}D+D\sqrt{A}$ for $D$ have a unique solution and is that solution symmetric? If you have specific questions about the efficiency of computing $D$ then you should be more specific about that and the context in which they could be answered. | |
| Mar 9, 2015 at 9:12 | comment | added | user68601 | @Carlo: as I expected the integral does not have an explicit expression. So I was wondering how it occurred that in the second link you came up with the answer $(A+B)^{1/2}=A^{1/2}(1+A^{-1}B/2)$? I mean which is the path that you follow to derive such a result? Because mine above has the disadvantage to provide an useless (at least for my purposes) solution. | |
| Mar 9, 2015 at 9:11 | comment | added | user68601 | @Igor: thank you for the reply! Your solution involves the eigen values and vectors of $A^{1/2}$, Am I wrong? If not, regrettably your solution is not feasible for my purposes as it does not allow me to develop further calculations. | |
| Mar 5, 2015 at 23:37 | comment | added | Igor Khavkine | Diagonalize, so that $\sqrt{A} = \mathrm{diag}(\sqrt{a_i})$. In the same basis, $B_{ij} = D_{ij} (\sqrt{a_i} + \sqrt{a_j})$, so that $D_{ij} = \frac{B_{ij}}{\sqrt{a_i} + \sqrt{a_j}}$, which is the unique solution and obviously symmetric. | |
| Mar 5, 2015 at 22:57 | comment | added | Carlo Beenakker | there is no closed-form expression for that integral, at least I do not know of any; given $A$ and $B$ you can evaluate it, but not in general. | |
| Mar 5, 2015 at 22:18 | comment | added | user68601 | Thank you Carlo again for the reply. What it is not clear to me is how from that integral we obtain $D=A^{-1/2}B/2$, or equivalently, the final expansion for the square root matrix $A^{1/2}+n^{-1/2}A^{-1/2}B/2$. | |
| Mar 5, 2015 at 20:24 | comment | added | Carlo Beenakker | the solution to $B=\sqrt A D+D\sqrt A$ is given in your second link: $D=\int_0^\infty e^{-t\sqrt{A}}Be^{-t\sqrt{A}}dt$; this is a symmetric matrix. | |
| Mar 5, 2015 at 19:01 | history | asked | user68601 | CC BY-SA 3.0 |