Timeline for Finding which members of a family of (possibly infinite-dimensional) matrices have trivial null space
Current License: CC BY-SA 4.0
11 events
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| Jun 21, 2021 at 20:07 | comment | added | A. Thomas Yerger | There are situations where you can think of such operators as "infinite matrices," I think one example is when you're dealing with e.g. $\ell^2(\mathbb{N})$? I think the point in such examples is that really you're just working with a separable Hilbert space, so you can get a countable basis and orthonormalize it. In fact I think once you've got this setup, the same proof as in finite dimensions will allow you to see that the operators with kernel 0 are open and dense. But I should not embarrass myself anymore and let someone who knows shed light on the functional analysis. | |
| Jun 21, 2021 at 10:50 | comment | added | exfret | Actually, I'm starting to double-guess myself about even my last comment (maybe you can correspond them?). If anything, I'll try to keep to MSE for questions in areas that I'm not too familiar with from now on even if they pop up in research. | |
| Jun 21, 2021 at 10:36 | comment | added | exfret | @AlfredYerger I'm starting to think this would indeed be better for MSE (despite occurring in research) since I have already made a huge beginner mistake - linear operators aren't in any sort of correspondence to any sort of set of infinite matrices! | |
| Jun 21, 2021 at 3:25 | history | edited | exfret | CC BY-SA 4.0 |
Clarification
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| Jun 21, 2021 at 3:14 | comment | added | exfret | @AlfredYerger You're right, I should be more specific. My hope would be that, in fact, no choice of $c$ works to make a nontrivial kernel. I'll add an extra section with some motivation to help clarify. | |
| Jun 21, 2021 at 2:48 | comment | added | A. Thomas Yerger | The finite dimensional case is "trivial" in the sense that for a given $c$ it is easy to hand to a computer and get the answer. What sort of a characterization are you looking for? | |
| Jun 21, 2021 at 1:07 | history | edited | exfret | CC BY-SA 4.0 |
edited body
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| Jun 21, 2021 at 1:05 | comment | added | exfret | In any case, the fact that S can be infinite should be enough to promote this to “research-level”, at least. | |
| Jun 21, 2021 at 1:03 | comment | added | exfret | Thanks for spotting the typo. Seems like “open and dense” should be the appropriate notion for “most” here yes. I don’t think we immediately get even this, though, and certainly characterizing the “good” c’s even in the finite dimensional case doesn’t strike me as a problem that is likely to be elementary enough for MSE. Perhaps I am wrong, though, and you see an easy line of attack to get some sort of useful result out of this. Btw, nice to see the >implying founder here (: | |
| Jun 21, 2021 at 0:46 | comment | added | A. Thomas Yerger | I'm not an expert on what might happen in the infinite dimensional case, but the finite dimensional case is really more appropriate for MSE at least. Also, matrices without kernel are open and dense in the collection of all $m \times n$ matrices, so this handles your question of what "most" should mean. Finally, I believe you have a typo in your question: there is no defintion of $l$ anywhere and yet it is a variable in your first big formula. | |
| Jun 20, 2021 at 21:49 | history | asked | exfret | CC BY-SA 4.0 |