Timeline for Product of matrices equal identity
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 30, 2021 at 15:33 | vote | accept | Apprentice | ||
| Nov 13, 2021 at 21:14 | answer | added | loup blanc | timeline score: 0 | |
| Oct 24, 2021 at 9:57 | comment | added | David Roberts♦ | Also, do you just a a solution, or all solutions? | |
| Oct 24, 2021 at 7:28 | comment | added | David Roberts♦ | 6) In fact, you might as well start with the Ansatz that $(PAP^\perp)^{-1} PSP^{-1} (PAP^\perp)^{-1} = diag(1,\ldots,1,-1,\ldots,-1)$, since you can get all possible solutions from this one by "undiagonalising". 7) Then you can make various assumptions about block structure of $P$, which amounts to choosing clever bases, and then the general case arises by inserting change of basis matrices. 8) you can probably assume that $A$ is actually positive definite, else a zero eigenvalue would break the equation. And so on. | |
| Oct 24, 2021 at 7:07 | comment | added | David Roberts♦ | Some random ideas: 1) you have a matrix squaring to the identity, all its eigenvalues are $\pm1$, and its determinant is $\pm1$ 2) you must have the inequality bound on $d\leq r$, else the equation has no solution (the various $PXP^\perp$ won't have full rank) 3) Is $P$ meant to diagonalise $S$, since $S$ is symmetric hence orthogonally diagonalisable? 4) you can get an equation $\det(PAP^\perp)^2 = \det(PSP^\perp)^4$. 5) I'd be inclined to break this into two equations, namely $X=(PAP^\perp)^{-1}$ and $(XPSP^\perp X)^2=I$. | |
| Oct 23, 2021 at 20:30 | comment | added | David Roberts♦ | I mean something more detailed, like what research problem does this come from? Trying to solve the scalar version is a small start, but not like what I was expecting. Like eg having a solid go at the 2x3 or 3x2 versions | |
| Oct 23, 2021 at 9:30 | comment | added | Apprentice | The context is the resolution of a system of matrix equations that leads to this equations. I have tried solving the scalar version of this equation. It leads to $p = sa^{-1}$. In the matrix case I don't know how to proceed. | |
| Oct 23, 2021 at 9:10 | review | Close votes | |||
| Nov 7, 2021 at 3:03 | |||||
| Oct 23, 2021 at 8:52 | comment | added | David Roberts♦ | What's the context for this question? What have you tried? | |
| Oct 23, 2021 at 6:24 | history | asked | Apprentice | CC BY-SA 4.0 |