Timeline for Expressing $-\operatorname{adj}(A)$ as a polynomial in $A$?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2010 at 2:38 | comment | added | Victor Protsak | Also, I forgot to mention: there is a natural (i.e. not relying on tricks and special cases) enumerative combinatorics proof of CH in Stanton-White's book. | |
| Jul 19, 2010 at 2:13 | comment | added | Victor Protsak | Quiochu: As in "going in circles" (i.e. first creating a difficulty by specialization/eigenvalues/... and then proudly resolving it with more unnecessary machinery). | |
| Jul 19, 2010 at 1:38 | comment | added | Bill Dubuque | @Q The equation in Victor's comment is indeed the equation that I started with, viz. $\rm d B = d C$ for $\rm d = det A$. | |
| Jul 19, 2010 at 0:39 | comment | added | Qiaochu Yuan | As in it doesn't work or as in you don't like it? In any case, maybe I should present this style of argument not as "the proof," but as a way to check such results. Many matrix identities are easier to verify for A with a special form, and I think it's good to know that there exist theorems saying that this is enough. Whether one chooses to use them is a matter of taste, I think. | |
| Jul 19, 2010 at 0:14 | comment | added | Victor Protsak | Qiaochu: Multiply your first displayed line by $\text{adj}(A),$ get $$\det A\ \text{adj}(A) = \det A(- p_1 I - p_2 A - \ldots - p_n A^{n-1})$$ and universally cancel $\det A$ per Bill's answer. The rest (specialization and density) is crud. | |
| Jul 18, 2010 at 19:11 | comment | added | Pete L. Clark | Sure, this is a nice proof. Depending upon taste, one might want to replace the "analytic" topology with the Zariski topology. The merit of this is that the Zariski density of the invertible matrices is even easier to establish: because $\mathbb{C}[x_1,\ldots,x_{n^2}]$ is a domain, its spectrum is an irreducible topological space, so every nonempty open subset -- like the set of all matrices with nonzero determinant -- is Z-dense. | |
| Jul 18, 2010 at 18:13 | history | answered | Qiaochu Yuan | CC BY-SA 2.5 |