Timeline for Bandwidth reduction of multiple matrices
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Mar 15, 2018 at 18:31 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
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| Mar 15, 2018 at 16:42 | review | Suggested edits | |||
| S Mar 15, 2018 at 18:31 | |||||
| Apr 7, 2012 at 20:33 | comment | added | Greg von Winckel | Additionally, it is possible to simultaneously diagonalize two matrices, but not with a similarity transformation. The matrix of eigenvectors to a symmetric definite matrix pencil $Ax=\lambda B x$ does diagonalize both $A$ and $B$, however, it is not an orthogonal matrix (see my first comment above). | |
| Apr 7, 2012 at 20:29 | comment | added | Greg von Winckel | Yes. That is exactly what I said. That is precisely why I was asking if there was a way to reduce them both to banded form. | |
| Apr 7, 2012 at 13:29 | comment | added | Chris Godsil | In general if $A$ is symmetric and $D$ is diagonal, it is not possible to simultaneously diagonalize them because this would imply that they commute. | |
| Apr 7, 2012 at 13:26 | comment | added | Greg von Winckel | Sorry, I should not have chosen the symbol $P$ in that case. $P$ is not a permutation matrix in my example, and $A$ is assumed to be full. As an example, suppose I have the generalized eigenvalue problems [ A V = V D_1 \Lambda_1 ] [ A U = U D_2 \Lambda_2 ] Now I have a matrice $V$ and $U$ such that the following matrices are diagonal: $V^\top A V$, $U^\top A U$, $V^\top D_1 V$, $U^\top D_2 U$ It seems, in general, impossible to find a single matrix that diagonalized $A$, $D_1$ and $D_2$, however, if it produced a banded matrix, that would still be useful for my purposes. | |
| Apr 7, 2012 at 8:25 | answer | added | Bart | timeline score: 1 | |
| Apr 6, 2012 at 16:35 | history | asked | Greg von Winckel | CC BY-SA 3.0 |