# Finding eigenvalues of $T + T^*$, where $T$ is the right-shift operator

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.

• what makes you think it has eigenvalues? – lcv Jul 8 '18 at 23:29
• actually, I wanted a proof showing that the operator does not have eigenvalues, and I did it. – Chilote Jul 9 '18 at 2:06
• 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). – Frederik vom Ende Mar 6 '19 at 18:08

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 $P_n(\lambda)$.
• I see. Thnx for letting me know. Well it seems we have to deal with a really cumbersome expression for $P_n$. – Arian Jul 8 '18 at 21:05
• $P_n(\lambda) = U_{n-1}(\lambda/2)$ 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. – Robert Israel Jul 9 '18 at 1:47