Hey, I can't seem to get anywhere with this question so any help you can provide would be much appreciated.
is a linear operator on a finite dimensional vector space , whereby equals the identity operator.
1. Prove that for any given vector u in , is the vector or an eigenvector with eigenvalue
2. Prove that V is indeed the direct sum of the eigenspaces and where is considered the set of eigenvectors with eigenvalue , together with .
I am really stuck here, so thank you for any help.
i think you need a bit more than that. you should, however, be able to write down the minimal polynomial for T, which will then tell you T cannot have any other eigenvalues.
i think it only fair to point out that for this result to be strictly true, we must require V to be over a field of char(F) ≠ 2 (in that case V(1) = V).
the minimal polynomial for a linear transformation T:V-->V, is the monic polynomial m(x) of least degree such that m(T) = 0 (the 0-transformation).
in this case, we have T^2 = I, so T^2 - I = 0, that is, T satisfies x^2 - 1 = 0.
now it could well be that T = I (there is nothing in your question that rules this out), or T = -I, in which case m(x) = x - 1, or m(x) = x + 1.
otherwise x^2 - 1 is the minimal polynomial for T, so in any case the only possible eigenvalues for T are in the set {-1,1}.
this shows V = V(1) + V(-1), but it does not prove that we have a direct sum, unless V(1)∩V(-1) = {0}. prove the intersection only has the 0-vector in it
(this is true, incidentally, for any two eigenspaces, can you see why?).