Define the operator $T:\ell^{2}(\mathbb Z)\to\ell^{2}(\mathbb Z)$ by $(T(x))_{n}=x_{n+1}+x_{n-1}$. Calculate the spectrum of $T$.
I am reviewing some functional analysis problems in preparation for an upcoming second course in functional analysis, and this one is eluding me. Any suggestions or tips?
I attempted to transform to an operator on $L^2(\mathbb R/\mathbb Z)$, but I have not been able to successfully manage that yet. I do still think such an approach, if successful, would make the spectrum calculation much easier.
Let $T(a_n):=(a_{n-1}+a_{n+1})=(a_1,a_0+a_2,a_1+a_3,\ldots)$.
Since $T=R+R^*$ (where $R$ is the right-shift operator and $R^*$ is the left shift), $T$ is self-adjoint and $\|T\|\le2\|R\|=2$, so the spectrum must lie in $[-2,2]$.
Part 1. Answer for $\ell^2(\mathbb{Z})$.
Since $R^*=R^{-1}$, then by the spectral mapping theorem $$\sigma(T)=\{\lambda+\lambda^{-1}:\lambda=e^{i\theta}\in\sigma(R)\}=[-2,2]$$
Part 2. Answer for $\ell^2(\mathbb{N})$
Checking for (real) eigenvalues: $T(a_n)=\lambda(a_n)$ gives the recurrence $$a_{n+1}=\lambda a_n-a_{n-1},\quad a_1=\lambda a_0,$$ This can be solved, either directly or using the theory of difference equations: $$(a_n)=a_0(1,\lambda,\lambda^2-\lambda,\lambda^3-2\lambda^2,\ldots)$$ $$a_n=\frac{a_0}{\alpha-\alpha^{-1}}\left(\alpha^{n+1}-\alpha^{-(n+1)}\right)=a_0\left(\frac{\sin(n+1)\theta}{\sin\theta}\right),$$ where $\alpha,\alpha^{-1}$ are the (complex) roots of $x^2-\lambda x+1=0$. It follows from $\lambda\in[-2,2]$ that $\alpha=e^{i\theta}$ and $\lambda=\alpha+\alpha^{-1}=2\cos\theta$.
One can check that there are no eigenvectors (in $\ell^2$) since $|\alpha|=1$, but there are approximate eigenvalues. Let $x_n:=(\sin(i+1)\theta)_{i\le n}$ but with $a_i=0$ for $i>n$. Then $$(T-\lambda)x_n=(0,\ldots,0,-\sin(n+2)\theta,\sin(n+1)\theta,0,\ldots)$$ $$\therefore \frac{\|(T-\lambda)x\|^2}{\|x\|^2}=\frac{\sin^2(n+2)\theta+\sin^2(n+1)\theta}{\sum_{i=0}^n\sin^2(i+1)\theta}\to0\mathrm{\ as\ }n\to\infty$$
Conclusion: The spectrum of $T$ is $[-2,2]$.