Timeline for Theorems for nothing (and the proofs for free)
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2009 at 14:03 | comment | added | Mark Meckes | Gil: maybe what was meant was the following: consider det(tI-A), and plug in A for t. Personally I wouldn't say this makes C-H "intuitively correct"; instead C-H is suggested by this simple heuristic. | |
| Nov 22, 2009 at 13:46 | comment | added | Steven Sivek | Then you just realize that det(tI-A) evaluated at A is some matrix whose entries are monstrously complicated polynomials in the n^2 entries of the matrix A, and since they're identically 0 on C^{n^2} each of those entries must be the zero polynomial; thus the theorem holds over any commutative ring as well. | |
| Nov 22, 2009 at 10:19 | comment | added | Ryan Budney | The cheezy-easy proof that works over the real or complex numbers is to observe diagonalizable matrices are dense in the space of matrices, and the theorem is true for diagonalizable matrices (by computation) then notice the set of matrices that satisfy the theorem are closed. If you want to avoid this kind of argument you can enhance your intuition with the Jordan Canonical Form. :) | |
| Nov 22, 2009 at 6:48 | comment | added | Gil Kalai | Indeed this is a wonderful theorem. Why is it intuitive correct? From all the first year algebra theorems it was the one where I had no intuition whatsoever. | |
| Nov 22, 2009 at 5:46 | history | answered | M.G. | CC BY-SA 2.5 |