Timeline for Recovering a Matrix After Multiplication By Its Transpose [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 27, 2012 at 10:10 | history | closed |
Will Jagy Yemon Choi Andrew Stacey S. Carnahan♦ |
too localized | |
| Feb 27, 2012 at 9:16 | comment | added | Federico Poloni | I really do not know where this habit of using $A^T$ or $A'$ where $A$ is symmetric comes from... | |
| Feb 27, 2012 at 8:10 | comment | added | Will Jagy | @Yemon, if I leave my mouse arrow resting on either unknown google, at the bottom of my browser it shows the user number, 1894 for one and 21740 for the OP. I am 3324 and you are 763, so 1894 started long after you but long before i did. Also, Gordon Pall used quaternions to parametrize $SO_3 \mathbb Q,$ from what I can see his recipe can always be rearranged by permutations to be a symmetric orthogonal matrix. I think his recipe is essentially the same thing they use for computer graphics now. | |
| Feb 27, 2012 at 7:37 | comment | added | Yemon Choi | To the OP: do you want all solutions, or just a possible solution? Are you perhaps also assuming P is positive? | |
| Feb 27, 2012 at 7:35 | comment | added | Yemon Choi | There seem to be two unknown (google)s in this discussion. Could at least one of you please take a moment to choose a pseudonym, or better still your actual name? | |
| Feb 27, 2012 at 7:35 | answer | added | Denis Serre | timeline score: 3 | |
| Feb 27, 2012 at 7:18 | comment | added | Chua KS | The converse of Will's observation also holds: ie. if $A$ is nonsingular, and $P=A_1^TA_1=A_2^TA_2$, then $R=A_2A_1^{-1}=R^T$ is orthogonal and $A_2=RA_1$. In this case if you think of the Euclidean lattice generated by the columns of $A$, then $P=A^TA$ is the Gram matrix (the quadratic form) and this just say that the Gram matrix determines the lattice upto a rotation. But this does not answer the question, after finding $P=A^TA$ by Cholesky, we know all solutions are given by $RA, R \in O(N)$, some of the $R$ will make $RA$ symmetric. | |
| Feb 27, 2012 at 6:37 | comment | added | Yemon Choi | There can be many different looking symmetric matrices, each of whose square is the identity matrix. | |
| Feb 27, 2012 at 6:19 | comment | added | user21740 | I apologize - the original question was formulated poorly. The matrix A is symmetric as well. | |
| Feb 27, 2012 at 6:15 | history | edited | user21740 | CC BY-SA 3.0 |
deleted 50 characters in body
|
| Feb 27, 2012 at 6:13 | comment | added | Will Jagy | You cannot estimate $A.$ Suppose $A$ is the Cholesky decomposition of $P,$ which is easy to compute and numerically stable. Then, given any orthogonal matrix $R,$ meaning $R^T \; R = I,$ then consider $B = R A.$ We get $$B^T \; B = (RA)^T \; R A = A^T \; R^T \; R \; A = A^T \; I \; A = P. $$ | |
| Feb 27, 2012 at 6:02 | history | edited | Andrés E. Caicedo | CC BY-SA 3.0 |
added 13 characters in body; edited title
|
| Feb 27, 2012 at 6:01 | comment | added | Yoav Kallus | Look up Cholesky decomposition | |
| Feb 27, 2012 at 5:57 | history | asked | user21740 | CC BY-SA 3.0 |