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Math Help - Prove T has infinitely many LI eigenvectors

  1. #1
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    Prove T has infinitely many LI eigenvectors

    Let X be an infinite-dimensional complex Banach space and T:X\to X a continuous/bounded linear operator. Suppose that for every nonzero e\in X, the "orbit space" \mathcal{O}_T(e):=\text{span}(T^ne)_{n=0}^\infty is finite-dimensional, say has dimension 1\leq d(e)<\infty.

    I would like to prove or disprove the following conjecture:

    Conjecture 1: T has infinitely many linearly independent eigenvectors.

    Conjecture 1 easily follows from the following:

    Conjecture 2: There is a sequence (e_n)_{n=0}^\infty of nonzero vectors in X such that \mathcal{O}_T(e_{N+1})\cap(\mathcal{O}_T(e_0)+ \cdots +\mathcal{O}_T(e_N))=\{0\} for all N.

    Proof that Conjecture 2 implies Conjecture 1: First we claim that for each nonzero e\in X, T has an eigenvector in \mathcal{O}_T(e). For proof, notice that since O_T(e) is T-invariant then the restriction T|_{\mathcal{O}_T(e)} of T to \mathcal{O}_T(e) is a linear operator on a finite-dimensional vector space. Being defined on a nonzero but finite-dimensional vector space, this means T|_{\mathcal{O}_T(e)} has eigenvalues. Thus we can find an eigenvector v\in\mathcal{O}_T(e) under T|_{\mathcal{O}_T(e)}, and the claim is proved. In particular, for each n we can find an eigenvector v_n\in\mathcal{O}_T(e_n). Next we claim that the resulting sequence (v_n)_{n=0}^\infty is linearly independent. For suppose otherwise, towards a contradiction. Then we can find a_0,\cdots,a_{N+1}\in\mathbb{C}, not all zero, such that 0=\sum_{n=0}^{N+1} a_nv_n. Without loss of generality suppose a_{N+1}=-1. Then v_{N+1}=\sum_{n=0}^Na_nv_n\in\mathcal{O}_T(e_0)+ \cdots +\mathcal{O}_T(e_N). However v_{N+1}\in\mathcal{O}_T(e_{N+1}), contradicting the hypothesis that \mathcal{O}_T(e_{N+1})\cap(\mathcal{O}_T(e_0)+ \cdots +\mathcal{O}_T(e_N))=\{0\}. \square

    So if anyone can help me prove either conjecture, that would be great. Thanks!
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  2. #2
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    Re: Prove T has infinitely many LI eigenvectors

    Hey hatsoff.

    Does the operator have to be Hermitian? (Forgive me, it's been a while since I looked at this stuff)?
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  3. #3
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    Re: Prove T has infinitely many LI eigenvectors

    Quote Originally Posted by chiro View Post
    Hey hatsoff.

    Does the operator have to be Hermitian? (Forgive me, it's been a while since I looked at this stuff)?
    Unfortunately no---I must prove this for a diverse class of operators on non-Hilbert Banach spaces. However I might be able to assume it has a finite spectrum. For what it's worth, I can also assume X is not reflexive. (The reflexive case doesn't really need this result.)
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  4. #4
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    Re: Prove T has infinitely many LI eigenvectors

    I don't think I can help you out on this one: This is a lot more general than the stuff I've been exposed to.
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