Prove T has infinitely many LI eigenvectors

**hatsoff** · February 3rd 2013, 08:53 AM

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!

**chiro** · February 3rd 2013, 11:03 PM

Hey hatsoff.

Does the operator have to be Hermitian? (Forgive me, it's been a while since I looked at this stuff)?

**hatsoff** · February 4th 2013, 04:19 AM

Originally Posted by chiro

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.)

**chiro** · February 4th 2013, 05:19 PM

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.

Thread: Prove T has infinitely many LI eigenvectors

LinkBack

Thread Tools

Display

Prove T has infinitely many LI eigenvectors

Re: Prove T has infinitely many LI eigenvectors

Re: Prove T has infinitely many LI eigenvectors

Re: Prove T has infinitely many LI eigenvectors

Similar Math Help Forum Discussions

Prove that Spec√2 contains infinitely many powers of 2.

[SOLVED] Prove there are infinitely many daffodil numbers

[SOLVED] Prove that there are infinitely many primes in the form 6k+5

Prove prime p divides infinitely many members of 1, 11, 111, 1111...

Prove that p divides infinitely many numbers...

Search Tags