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.