Timeline for Slick proof?: A vector space has the same dimension as its dual if and only if it is finite dimensional
Current License: CC BY-SA 4.0
9 events
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| Jan 31, 2023 at 20:33 | comment | added | PatrickR | Yes, you are right. I mistakenly thought that if all $\alpha^p=\alpha$, the p-th row will be the same as the 1st row (starting to count at row 0). But in that case all these $\alpha$ must be in the prime subfield and there won't be a p-th row. | |
| Jan 31, 2023 at 12:57 | comment | added | Alcides Buss | @PatrickR, I didn't understand your comment. The Vandermonde matrix is always invertible, over any field, despite of its characteristic. | |
| Jan 21, 2023 at 6:38 | comment | added | PatrickR | Nice proof when the field has characteristic $0$. But may need adjustment in characteristic $p$, as $\alpha^p$ could be the same as $\alpha$ and the Vandermonde matrix may not be invertible. | |
| Apr 18, 2022 at 14:16 | comment | added | Alcides Buss | I meant $\dim(V^*)\geq |k|$ above, proving this inequality was the main point of my argument. And then it follows $\dim(V^*)=|V^*|$. | |
| Apr 17, 2022 at 23:11 | comment | added | Alcides Buss | @Ryan Budney, I didn't pay too much attention to Pace Nielsen's argument in the first place. They are based partially on the same ideas. My point was too give a possibly elementary proof of the fact that $|V^*|\geq |k|$. Pace Nielsen adds at the end a comment that the argument does not work for fields of large cardinality. And my argument actually works for every field, of arbitrary cardinality. | |
| Apr 16, 2022 at 19:09 | comment | added | Ryan Budney | This is a restatement of Pace Nielsen's argument. | |
| Apr 16, 2022 at 18:14 | history | edited | Alcides Buss | CC BY-SA 4.0 |
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| S Apr 15, 2022 at 21:19 | history | answered | Alcides Buss | CC BY-SA 4.0 | |
| S Apr 15, 2022 at 21:19 | history | made wiki | Post Made Community Wiki by Alcides Buss |