Jacobi operator's spectrum in $l_2$ 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.
 A: 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 $\lambda \in \sigma (S+S^*)$.
