Timeline for On $XX'=I$ such that $AX=XB$ is true
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2015 at 7:09 | comment | added | Joshua Grochow | @RobertIsrael: Sorry, of course I was being silly. I was thinking "diagonalizable", not normal. But I think the point of my previous comment is still valid: the normal case (or even the diagonalizable case) seems to miss some of the essential difficulties of the problem. | |
| Dec 8, 2015 at 6:57 | comment | added | Robert Israel | @JoshuaGrochow Normal matrices are dense? Certainly not: they form a closed set, since they are the solutions of an equation of the form $F(X) = 0$ with $F$ continuous. | |
| Dec 7, 2015 at 5:17 | comment | added | Joshua Grochow | Although normal matrices are dense, the real difficulty in this problem stems from non-normal matrices (as your answer shows). I don't think there is any such simple criterion for general matrices... | |
| Dec 7, 2015 at 1:27 | comment | added | Turbo | Ok how about (2)? | |
| Dec 7, 2015 at 1:14 | history | answered | Robert Israel | CC BY-SA 3.0 |