Timeline for answer to Precise meaning of "picking a basis"? by Iosif Pinelis
Current License: CC BY-SA 4.0
Post Revisions
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3 at 19:02 | comment | added | Iosif Pinelis | @TimothyChow : Even the first version of the answer presented a basis-free definition of the trace, which later turned out to be significantly shorter and more elementary than the Wikipedia definition. I don't think that this qualifies as nothing substantive. That I was not aware of the Wikipedia definition when posting mine does not make my definition less substantive. Also, in 20 hours after the first version of my answer I added a basis-free definition of finite dimensionality. | |
| Feb 3 at 13:46 | comment | added | Timothy Chow | @M.Winter Probably a lot of the downvotes were for the original version of the answer; as you can see from Iosif's comment, he didn't bother reading what Buzzard wrote, and so didn't contribute anything substantive on his first attempt. | |
| Jan 18 at 3:26 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 104 characters in body
|
| Jan 16 at 23:32 | comment | added | M. Winter | I don't get all the downvotes on this answer. Yes, the give definition might be equivalent to the blog post. But the important point (that I have not seen raised in any other answer so far) is that if we define finite-dimensional in the right way then we get a base free construction of the trace. It would be interesting to see how much of linear algebra can be built on the existence of a trace alone, without choosing a basis. | |
| Jan 16 at 15:58 | comment | added | Iosif Pinelis | @EmilJeřábek : Thank you for your further comments. | |
| Jan 16 at 15:49 | comment | added | Emil Jeřábek | Hmm. Though this likely depends on the axiom of choice as well. Without choice, it may well turn out that there is no nonzero linear functional $\ell\colon V\to F$ at all. Never mind. | |
| Jan 16 at 15:46 | comment | added | Emil Jeřábek | The existence and nonuniqueness of $T$ heavily relies on the axiom of choice. I think a better basis-free definition of finite dimensionality could be that $V$ is finitely dimensional iff $L(V)$ is spanned by $\{A_{\ell,v}:v\in V,\ell\colon V\to F\}$. | |
| Jan 16 at 15:31 | comment | added | Iosif Pinelis | @EmilJeřábek : Thank you for your comment. This should now be fixed. | |
| Jan 16 at 15:31 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 322 characters in body
|
| Jan 16 at 15:02 | comment | added | Emil Jeřábek | I don't think the claims are correct. It is the other way round: if $V$ is infinite-dimensional, a trace satisfying Definition 1 exists, but it is not unique. Let $W\subseteq L(V)$ be the subspace of maps with finite-dimensional image, and using the axiom of choice, let $W'$ be its direct complement. Then the recipe in Def. 1 defines a unique operator on $W$, which we may extend to all of $L(V)$ by picking an arbitrary operator on $W'$. | |
| Jan 16 at 14:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
deleted 135 characters in body
|
| Jan 16 at 13:48 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 22 characters in body
|
| Jan 16 at 13:40 | comment | added | Iosif Pinelis | @WillSawin : Thank you for your comment. My definition is indeed equivalent to one in the blog post (which I had hardly read). However, as it is now discussed in the answer, my definition is stated in more elementary terms and is shorter and more direct. I have now also amplified the discussion on the existence of a trace operator and added a discussion on whether it is possible to formalize what it means for a proof to pick a basis as far as the notion of the trace is concerned. | |
| Jan 16 at 13:32 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 2102 characters in body
|
| Jan 15 at 19:28 | comment | added | Will Sawin | This definition is essentially given in the linked blog post. The point is then raised that the proof that such a linear map is unique relies on the existence of a basis. | |
| Jan 15 at 19:28 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 528 characters in body
|
| Jan 15 at 19:17 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |