# 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.

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 $$\lambda \in \sigma (S+S^*)$$.

• Thank you very much for clear explanation! I'm interesting can the last part pe proved by more simple methods excluding Toeplitz algebra? Commented Nov 5, 2020 at 19:55
• 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.
– Ruy
Commented Nov 7, 2020 at 18:14