- ..., 4, 14, 23, 34, 42, 50, __, 72, 81, ...
- ..., a, f, p, n, __, m, ...
- ..., 1, 1, 1, 1, 1, __, ....
Classifying Critical Points
1 hour ago
Aspiring mathematician Daniel McLaury writes about mathematics and related topics. Content level ranges from the elementary on upwards.
Proof: We know that C fixes V. Let W be a subspace of V of minimal positive order which is fixed by C. We claim that each vector in W is an eigenvector of each element of C. Suppose not; then there is some counterexample Mw≠λw. M fixes W though, so the restriction of M to W is a complex linear map in its own right; in particular, it has to have some eigenvector in W, say Mx=μx. Let's have Wμ denote the μ-eigenspace of M restricted to W. Then the previous two sentences in effect say that that 0 <>μ <>μ is fixed by C -- take any w in Wμ and any N in C and notice that M(Nw) = (MN)w = (NM)w = N(μw) = μ(Nw), i.e. Nw is a μ-eigenvector of M, and since N, being after all from C, fixes W, it's an element of Wμ. But wait! Before we said that W was as small as you could get and still be fixed by C, but we've just shown that Wμis both fixed by C and smaller than W. This can't be, so our assumption must have been wrong -- W must consist exclusively of eigenvectors for every element of C.Quick question: how many different eigenvalues can be associated with the subspace W in this proof, for any given transformation? For all of them?