Does the operator have to be Hermitian? (Forgive me, it's been a while since I looked at this stuff)?
Let be an infinite-dimensional complex Banach space and a continuous/bounded linear operator. Suppose that for every nonzero , the "orbit space" is finite-dimensional, say has dimension .
I would like to prove or disprove the following conjecture:
Conjecture 1: has infinitely many linearly independent eigenvectors.
Conjecture 1 easily follows from the following:
Conjecture 2: There is a sequence of nonzero vectors in such that for all .
Proof that Conjecture 2 implies Conjecture 1: First we claim that for each nonzero , has an eigenvector in . For proof, notice that since is -invariant then the restriction of to is a linear operator on a finite-dimensional vector space. Being defined on a nonzero but finite-dimensional vector space, this means has eigenvalues. Thus we can find an eigenvector under , and the claim is proved. In particular, for each we can find an eigenvector . Next we claim that the resulting sequence is linearly independent. For suppose otherwise, towards a contradiction. Then we can find , not all zero, such that . Without loss of generality suppose . Then . However , contradicting the hypothesis that .
So if anyone can help me prove either conjecture, that would be great. Thanks!