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I want to find the spectrum (with points classification) of operator $A$ in $l_2$, acting on the standart basis $\{e_n\}$ in the following way $$ Ae_1 = ae_1 + be_2, \ Ae_n = be_{n-1} + ae_n + be_{n+1}, \ n\geq 2 $$ Of course we can assume that $b\neq 0$, since on the other hand the problem is simple.

My attempts. First of all I tried to find point spectrum $\sigma_p(A) =\{\lambda \in \mathbb{C}: \ker(A - \lambda I) \neq \{0\} \}$, where $I$ is an identity operator. Let $x = (x_1,x_2, \ldots)\in l_2$. We obtain equations of the form $$ Ax = \lambda x \Leftrightarrow \begin{cases} x_2 = \frac{(\lambda -a)x_1}{b} \\ x_3 = \frac{(\lambda -a)x_2}{b}-x_1 \\ x_4 = \frac{(\lambda -a)x_3}{b}-x_2 \\ \ldots \\ x_n = \frac{(\lambda -a)x_n-1}{b}-x_{n-2} \\ \ldots \end{cases} $$ Also we can obtain the equations for $x_n$ in the form $$ x_n = p_n\left(\frac{\lambda -a}{b}\right)x_1 $$ where $p_n(x)$ is a polynomial of degree $n-1$. But the form of polynomials is remains unclear. Also this sequnce $x$ should belongs to $l_2$, that is $$ \sum_{n\geq 1}|x_n|^2 \leq \infty $$ It is clear that for $\lambda = a$ we can construct such a sequence, so $$ a \in \sigma_p(A) $$ But what can we say after that? Also I found that this operator has tha following propery $$ A^* = \overline{A} $$ in particular it is normal operator.

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Let $S$ be the right shift operator on $\ell^2$ given by $S(e_n)=e_{n+1}$. Observing that $$ A=aI+b(S+S^*), $$ it is enough to compute the spectrum of $S+S^*$ since one then has that $$ \sigma (A) = a + b\sigma (S+S^*), $$ by the spectral mapping Theorem. The classification of spectral elements will also follow because the class of any spectral value $\lambda \in \sigma (S+S^*)$ will be the same as the class of $a+b\lambda $, as a spectral value of $A$.

Observing that $S+S^*$ is a self-adjoint operator with norm no bigger than $2$, we see that $\sigma (S+S^*)\subseteq [-2, 2]$.

Speaking of eigenvalues, suppose that $\lambda $ lies in the point spectrum of $S+S^*$, and let $x=(x_n)_{n=1}^\infty $ be an eigenvector. Then $x$ satisfies the difference equation $$ x_{n+1}+x_{n-1} = \lambda x_n, $$ or, equivalently $$ x_{n+2} - \lambda x_{n+1} +x_{n} = 0, $$ whose characteristic polynomial is $$ z^2-\lambda z+1 = 0. $$ So the characteristic roots are $$ z={\lambda \pm \sqrt{\lambda ^2-4}\over 2 } $$ $$ ={\lambda \pm i\sqrt{4-\lambda ^2}\over 2 }. $$ Assuming that $\lambda \in [-2, 2]$, we see that the characteristic roots have absolute value 1, so the solutions $x_n$ do not converge to zero and hence cannot belong to $\ell^2$. In other words, there are no eigenvalues and hence the point spectrum of $S+S^*$ is empty.

Since $S+S^*$ is self-adjoint, it follows that its spectrum is then the same as the continuous spectrum.

The closed *-algebra $\mathcal T$ of operators on $\ell^2$ generated by $S$ is called the Toeplitz algebra. It is well known that $\mathcal T$ contains the algebra $\mathcal K$ formed by all compact operators and that the quotient $\mathcal T/\mathcal K$ is isomorphic to $C(S^1)$, namely the algebra of all continuous, complex valued functions on the unit circle $S^1$.

The image of $S$ under the quotient map $$ \pi :\mathcal T \to \mathcal T/\mathcal K = C(S^1) $$ is known to be the identity function $$ f(z)=z,\quad \forall z\in S^1, $$ so the image of $S+S^*$ is the function $$ g(z) = f(z)+\overline{f(z)} = 2\Re(z). $$

Since homomorphisms shrink spectra, we conclude that $$ \sigma (S+S^*) \supseteq \sigma (\pi (S+S^*)) = \sigma (g) = \text{Range}(g)=[-2,2], $$ so we finally get $$ \sigma (S+S^*) = \sigma _c(S+S^*) = [-2,2], $$ whence $$ \sigma (A) = \sigma _c(A) = [a-2b,a+2b], $$


EDIT: Here is an elementary proof, not using the Toeplitz algebra, that $[-2, 2]\subseteq \sigma (S+S^*)$.

Recall that the search for eigenvalues for $S+S^*$ leads us to consider the initial value problem $$ \left\{ \matrix{x_{n+2} - \lambda x_{n+1} +x_{n} = 0, \cr x_2 = \lambda x_1, } \right. \tag 1 $$ whose characteristic polynomial is $$ z^2-\lambda z+1 = 0. $$ Under the assumption that $\lambda \in [-2, 2]$, the characteristic roots are the two conjugate complex numbers $$ z ={\lambda \pm i\sqrt{4-\lambda ^2}\over 2 }, $$ both of which have absolute value is $1$, and hence may be expressed as $z=e^{\pm i\theta }$, with $\theta \in [0,\pi ]$.

According to the Wikipedia entry for "Linear difference equation" (https://en.wikipedia.org/wiki/Linear_difference_equation), in the section on "Converting complex solution to trigonometric form", the solutions have the form $$ x_n = K\cos(n\theta+\psi ), %{2{\sqrt {\gamma ^{2}+\delta ^{2}}}\cos(n\theta+\psi )}, $$ where $K$ and $\psi $ are constants.

Fixing any nonzero solution $x = (x_n)_n$, notice that when $\theta $ is a rational multiple of $2\pi $, the $x_n$ are periodic. Otherwise the $x_n$ describe a dense set in some symmetric interval. In any case the $x_n$ fail to converge to zero and in particular $$ \sum_{n=1}^\infty |x_n|^2 = \infty , $$ so $x$ does not belong to $\ell ^2$. Incidentally this is why $S+S^*$ admits no eigenvalues. Nevertheless, the existence of nonzero solutions to (1) will be our main tool in showing that every $\lambda $ in $[-2,2]$ belongs to the spectrum of $S+S^*$.

In order to prove this, fix any $\lambda \in [-2, 2]$, and any nonzero solution $x = (x_n)_n$ to (1). For each $k\geq 1$, let $$ x^k = (x_1,x_2,\ldots ,x_k,0,0\ldots ), $$ keeping in mind that $$ \lim_{k\to \infty }\|x^k\|=\infty . \tag 2 $$ We then have that $$ (S+S^*)(x^k)-\lambda x^k = $$ $$ \matrix{ =&&(&0,& x_1,&x_2,&\ldots ,&x_{k-2},&x_{k-1},&x_k,&0,&\ldots &)\cr &+&(&x_2,&x_3,&x_4,&\ldots ,&x_k,&0,&0,&0,&\ldots &) \cr &-&(&\lambda x_1,&\lambda x_2,&\lambda x_3,&\ldots ,&\lambda x_{k-1},&\lambda x_k,&0,&0,&\ldots &)& =\cr =&&(&0, &0, &0, &\ldots , &0, &x_{k-1}-\lambda x_k, &x_k, &0,&\ldots &).&}. $$ Observing that $|x_n|\leq K$, for every $n$, we then see that $$ \|(S+S^*)(x^k)-\lambda x^k\| \leq |x_{k-1}| + |\lambda x_k| + |x_k| \leq 2K+|\lambda |K. $$ From (2) we then deduce that $S+S^*-\lambda I$ sends arbitrarily large vectors (the $x^k$) to vectors of bounded size, so this shows that $S+S^*-\lambda I$ is not invertible and hence that $\lambda \in \sigma (S+S^*)$.

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    $\begingroup$ Thank you very much for clear explanation! I'm interesting can the last part pe proved by more simple methods excluding Toeplitz algebra? $\endgroup$ Commented Nov 5, 2020 at 19:55
  • $\begingroup$ Let me know it the above elementary proof is OK. Nevertheless I strongly suggest that you study the Toeplitz algebra since it gives a very good perspective to this and many other problems in operator theory. $\endgroup$
    – Ruy
    Commented Nov 7, 2020 at 18:14

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