1
$\begingroup$

I have a question regarding operators and I found an answer but I do not want to type the full answer. If you want to do it I can give you the pieces of the answer (see below). Once the first well-written answer is posted I will accept it. (The answer below is incomplete.)

Let $T:\ell^2\to\ell^2$ be the right-shift operator $T(x_1,x_2,x_3,\dots)=(0,x_1,x_2,\dots)$. Then its adjoint is the left-shift operator $T^*(x_1,x_2,x_3,\dots)=(x_2,x_3,x_4,\dots)$.

Question: How to prove that the self-adjoint operator $T+T^*$ does not have eigenvalues? $$(T+T^*)(x_1,x_2,x_3,\dots)=(x_2,x_1+x_3,x_2+x_4,\dots,x_n+x_{n+2},\dots)$$

Answer in Pieces: First, suppose that $x=(x_n)_n$ is an eigenvector of $T+T^*$ corresponding to some eigenvalue $a\in\mathbb{C}$. In order to find the closed form of $x_n$ and the justification of its uniqueness see this question (click here).

Second, to prove that the sequence $(x_n)_n$ is not an eigenvector it is enough to prove that $\lim x_n\neq0$ see the hint in this question.

Finally, the final piece is in the comments and answer of this question.

$\endgroup$
  • $\begingroup$ what makes you think it has eigenvalues? $\endgroup$ – lcv Jul 8 '18 at 23:29
  • $\begingroup$ actually, I wanted a proof showing that the operator does not have eigenvalues, and I did it. $\endgroup$ – Chilote Jul 9 '18 at 2:06
  • $\begingroup$ Only slightly related but one can give a more simple argument that $T+T^*$ does not have any eigenvalues (which circumvents having to deal with recurrence relations). $\endgroup$ – Frederik vom Ende Mar 6 '19 at 18:08
2
$\begingroup$

Suppose that $x:=(x_1,x_2,...)$ is an eigenvector of the operator $T+T^*$ and $\lambda$ its respective eigenvalue (we are assuming that $\sigma(T+T^*)\neq \emptyset$ otherwise we have nothing to say). From the equation $$(T+T^*)x=\lambda x$$ follow the following equations $$x_2=\lambda x_1,x_1+x_3=\lambda x_2,...,x_n+x_{n+2}=\lambda x_{n+1},....$$ These imply $$x_2=\lambda x_1, x_3=(\lambda^2-1)x_1,x_4=(\lambda^3-2\lambda)x_1,..., x_n=P_n(\lambda)x_1,...$$ where $P_n(\lambda)$ is a polynomial in $\lambda$ of degree $n-1$ satisfying the recursive relation $$P_n(\lambda)=\lambda P_{n-1}(\lambda)-P_{n-2}(\lambda)$$ for $n\geqslant 2$. With convention let $P_{0}(\lambda)=0$ and $P_1(\lambda)=1$. Since $x\in l^2(\mathbb{N})$ then $$||x||^2_{l^2}=|x_1|^2+|x_2|^2+...+|x_n|^2+...=|x_1|^2(|P_1(\lambda)|^2+|P_2(\lambda)|^2+...+|P_n(\lambda)|^2+...)<\infty$$ It is desirable to get a formula for $P_n(\lambda)$ and then getting necessary conditions on $\lambda$ so that the infinite sum above is finite. But we have a recursive relation of second degree together with two initial conditions. So in principle you should be able to get a formula for $P_n(\lambda)$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Hi @Arian, thanks for your answer but of course this is something that I also can do. As it is now, I cannot consider it as a full answer. Actually, it is what I did: see math.stackexchange.com/q/2844980 $\endgroup$ – Chilote Jul 8 '18 at 21:00
  • $\begingroup$ I see. Thnx for letting me know. Well it seems we have to deal with a really cumbersome expression for $P_n$. $\endgroup$ – Arian Jul 8 '18 at 21:05
  • 1
    $\begingroup$ @Chilote You are just asking to check that there are no solutions to a second order linear recurrence with constant coefficients except for linear combinations of exponentials? It is simple, just as in ODE: construct a solution to a given "initial value problem" using the exponentials, and then show that any two solutions to that initial value problem must be equal (by subtracting them and checking that they solve the initial value problem with zero initial data). An equivalent proof is to use linear algebra and the Jordan normal form. $\endgroup$ – Ian Jul 8 '18 at 21:08
  • 1
    $\begingroup$ $P_n(\lambda) = U_{n-1}(\lambda/2)$ where $U_n$ are the Chebyshev polynomials of the second kind. But I'm pretty sure there won't be any $\lambda$ for which $\sum_n P_n(\lambda)^2$ converges. $\endgroup$ – Robert Israel Jul 9 '18 at 1:47
  • 1
    $\begingroup$ The final piece is in the comments of the question and the answer of math.stackexchange.com/q/2845126 $\endgroup$ – Chilote Jul 9 '18 at 2:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.