Timeline for Collection of matrices in $SU_{\mathbb{C}}(n)$ with given family of eigenvectors
Current License: CC BY-SA 3.0
10 events
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| Aug 31, 2017 at 23:14 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 1, 2017 at 23:04 | comment | added | Henry.L | @ChristianRemling Let me sit down and give it a second thought but thanks a lot!! | |
| Aug 1, 2017 at 23:03 | comment | added | Christian Remling | I'm not sure a highbrow view is helpful here when there's the simple answer "exactly those $N$ that take the standard basis to a basis of ev's of $M$," and then you want only those that are also in $SO(n)$, or at least that's how I would approach it. | |
| Aug 1, 2017 at 22:56 | answer | added | paul garrett | timeline score: 1 | |
| Aug 1, 2017 at 22:56 | comment | added | Henry.L | @ChristianRemling So there is not a specific way of characterizing all these $N$ besides checking if they actually fall in $SO(n)$? | |
| Aug 1, 2017 at 22:50 | comment | added | Christian Remling | If $x_j$ are the normalized ev's of $M$, in any order (and written as columns), then you must take $N=(e^{i\alpha_1}x_1,\ldots, e^{i\alpha_n}x_n)$. So the final answer is: those $N$'s of this form that happen to be in $SO(n)$ (if any). If $M$ has multiple eigenvalues, this is still the answer, but you have more choices now. | |
| Aug 1, 2017 at 22:46 | comment | added | Christian Remling | It's still not quite right, I think. For example, the $N$'s won't normally form a subgroup; $N=1$ typically won't work. | |
| Aug 1, 2017 at 22:25 | history | edited | Henry.L | CC BY-SA 3.0 |
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| Aug 1, 2017 at 14:11 | history | edited | YCor |
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| Aug 1, 2017 at 13:21 | history | asked | Henry.L | CC BY-SA 3.0 |