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Define on operator $T$ on the space of double sided sequences $$\ell_2 \left( \mathbb{Z} \right) = \{ \{ a_n \}_{n = - \infty}^{\infty} \ ; \sum_{n= -\infty} ^{\infty} \left| a_n \right|^2 < \infty \}$$ as following $Tx = y$ so that $y_n =-x_{n-1}+x_n - x_{n+1}$. Find $T$'s spectrum.

I really don't know how to approach this... What I tried so far, is finding eigenvalues and I didn't succeed. Obviously $T = I - S_l - S_r$ where $S_l, S_r$ are left and right shifts, but what else can I draw from this regarding the spectrum?

EDIT

Since $T = I - S_l - S_r$, the matrix representation of $T$ is $$T_k^j = \cases{ 1 \qquad ,k=j \\ -1 \qquad ,k-j \in \mp1 }$$ so it's easy to see that $T$ is symmetric thus $T^*=T$... Is this even related?

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The space $\ell^2(\mathbb Z)$ is naturally isomorphic with $L^2(\mathbb T)$, via the identification $\delta_n\longmapsto (t\mapsto e^{in\pi\,t})$. Under this identification, the shifts are mapped respectively to the multiplication operators $M_z$ and $M_{z^{-1}}$. So the spectrum of $S_l+S_r$ is the same as that of $M_{z+z^{-1}}$.

For a multiplication operator, its spectrum is the closure of the range. So $$ \sigma(S_r+S_l)=\overline{\{z+z^{-1}:\ z\in\mathbb T\}} =\overline{\{z+\bar z:\ |z|=1\}}=\overline{\{2\text{Re}\,z:\ |z|=1\}}=[-2,2]. $$ Thus $$ \sigma(I-S_r-S_l)=\{1-t:\ t\in[-2,2]\}=[-1,3]. $$

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    Can we say anything similar about the spectrum $\sigma(S_r+S_l)$ on $\ell^2(\mathbb{N})$? As far as I can tell, we no longer have a nice natural isomorphism with $L^2(\mathbb{T})$, and I'm wondering how the spectrum changes. – 2017-04-30
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    The spectrum is still $[-2,2]$, but the argument above does not work. – 2017-05-01
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    It is possible to do it more or less by hand, or by using that the unilateral shift is the Toeplitz operator with symbol $z+z^{-1}$; now the range $[-2,2]$ of the function is the essential spectrum, but as the norm is at most $2$, there is no room left for anything else. – 2017-05-01