boot camp day 11

today so far has been up and down. i fluctuate between feeling really confident and good about the upcoming test and feeling like i don't know anything and am going to fail. mostly how i feel depends on how a particular problem or lecture is going... but anyway, i've been working during lunch and after class with a kid in the program, trying to knock out as many problems as we can, and that's been good. in analysis, more series stuff, and then we started riemann integration.

in linear algebra, the bulk of the class was spent solving old basic exam problems, which was good. at the beginning, though, he talked a bit about the cayley hamilton theorem, which says that a matrix satisfies its own characteristic polynomial. his way to approach this problem was as follows (my discussion here is a rough sketch of what's happening): it is fairly trivial to show that this result holds for a matrix with n distinct eigenvalues (where n is the dimension of the vector space). but then, you can show that the set of matrices with n distinct eigenvalues is dense in the set of all n by n matrices. then, by using continuity of the determinant function (which is just a polynomial) you can argue that the result extends to arbitrary matrices. here's how he went about showing that the set of matrices with n distinct eigenvalues is dense (this is the cool part): given a matrix A, we can take its characteristic polynomial P, which is a polynomial depending on the entries of A. there is also something called the "discriminant" function, say D, which is a function of polynomials (D=D(P)) and is itself a polynomial in the coefficients of P. this function has the property that D(P)=0 if and only if P does -not- have n distinct eigenvalues. now, since there ARE matrices with n distinct eigenvalues, we know that the discriminant function D is not identically 0. on the other hand, it is a general property of a polynomial that if it is 0 on some open set, then it is identically 0. so, if we look at the set of all polynomials P such that D(P)=0, that is, the set of all polynomials with fewer than n distinct eigenvalues, then this set cannot have any interior! (for D is a polynomial; so if this set had an interior, D would be 0 on an open set, and therefore identically 0, which we know it is not, by the existence of matrices with n distinct eigenvalues) thus, since {P such that P(D)=0} has no interior, its complement is dense in the space. its complement is exactly those matrices with n distinct eigenvalues. anyway, he also proved that the discriminant D(P) exists and is a polynomial in the coefficients of P.

i think i will eat tuna and spinach salad for dinner and listen to the cardinals game now. thanks for reading!