Timeline for Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?
Current License: CC BY-SA 2.5
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 26, 2010 at 19:34 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 87 characters in body
|
| Jul 19, 2010 at 2:21 | comment | added | Victor Protsak | Bill, we are in a complete agreement about your last point. My explanation of the ubiquity of the phenomenon of willingly adding complexity is captured by the proverb "If you are a hammer then everything is a nail". | |
| Jul 18, 2010 at 22:36 | comment | added | Bill Dubuque | @Victor: to ensure there is no confusion, I remark that my use of "generic" above is not intended to denote anything topological or geometrical. Rather, it is meant to be understood as exploiting the universality of a free objects. The proof I gave does not require any knowledge of topology or (algebraic) geometry. I'm not saying that imposing other such viewpoints isn't interesting or useful - just that such is not required for problems of this sort. Further doing so adds complexity to what is - at the heart - trivial (yet elegant) algebra. | |
| Jul 18, 2010 at 19:00 | history | rollback | Bill Dubuque |
Rollback to Revision 14
|
|
| Jul 18, 2010 at 19:00 | history | rollback | Bill Dubuque |
Rollback to Revision 13
|
|
| Jul 18, 2010 at 18:38 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 4 characters in body
|
| Jul 18, 2010 at 18:22 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 38 characters in body; deleted 1 characters in body; added 5 characters in body
|
| Jul 18, 2010 at 17:51 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 28 characters in body; Post Made Community Wiki
|
| Jul 18, 2010 at 17:34 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 30 characters in body
|
| Jul 18, 2010 at 17:27 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 2 characters in body
|
| Jul 18, 2010 at 17:26 | history | rollback | Bill Dubuque |
Rollback to Revision 7
|
|
| Jul 18, 2010 at 16:12 | comment | added | Victor Protsak | Here is a question that you might answer. As you say, CH is widely known, but what about the OP's identity? I have a feeling that this is something that once upon a time was equally standard (early 20th century, when Kronecker-Weierstrass-Frobenius methods were widely employed), but I can't remember seeing it stated explicitly in any recent texts. Do you know the history of this identity? When did this transition from "polynomial universality" to "eigenvalues and density" approach (also evident in the current treatment of Jordan normal form without developing elementary divisors) occur? | |
| Jul 18, 2010 at 16:04 | comment | added | Victor Protsak | "Density argument" is ambiguous: I interpreted "the invertible case is already enough" in the first comment as "$\{A:\det A\ne 0\}$ is a Zariski dense subset of the affine space of $n\times n$ matrices" (over $C$, as it were), which is exactly your "working generically allows us to cancel $d$". As you can see, I am not happy with handwaving in such matters: I often found that it masks incomplete understanding (cf the link I gave above to discussion of what constitutes a proof of CH itself). This is typically manifested in starting an answer with "Hint" :) | |
| Jul 18, 2010 at 15:34 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
Tiny display of pedancy
|
| Jul 18, 2010 at 15:25 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 3 characters in body
|
| Jul 18, 2010 at 15:25 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
Tiny display of pendancy
|
| Jul 18, 2010 at 15:16 | comment | added | Bill Dubuque | @Victor: but Quiaochu's hint employs a density argument, not polynomial universality - which is the essence of the matter in my approach. Yes, I presume Cayley-Hamilton for commutative rings but this is so widely known it is even in Wikipedia, besides Jacobson BA1, etc. (not to mention Nakayama inspired generalizations, e.g. Atiyah & Macdonald) | |
| Jul 18, 2010 at 15:11 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 16 characters in body; added 2 characters in body
|
| Jul 18, 2010 at 14:54 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 21 characters in body; deleted 1 characters in body
|
| Jul 18, 2010 at 14:47 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 43 characters in body
|
| Jul 18, 2010 at 14:37 | history | edited | Bill Dubuque | CC BY-SA 2.5 |
added 1594 characters in body
|
| Jul 18, 2010 at 8:30 | comment | added | Victor Protsak | Yes, this is probably what Quiaochu had in mind. A couple of points: (1) There is some work left in showing that Cayley-Hamilton holds universally, which some books fail to emphasize (hence many people are unaware how to prove it, in line with your last comment); and (2) The usual proof of CH relies on factorization $\det(A-\lambda)=(A-\lambda)\operatorname{adj}(A-\lambda)$ and this is needed again in the "black box" proof you've presented in the form $\det A=A\operatorname{adj} A$ (hence after unwinding the argument, it ends up being applied twice). That's why I prefer a direct argument. | |
| Jul 18, 2010 at 6:42 | history | answered | Bill Dubuque | CC BY-SA 2.5 |