Spectrum of operator $(Tx)_{i}=\sum_{j=0}^{n}\alpha_{j}x_{i+j}$

Question

Note: Please do not give a solution; I would prefer guidance to help me complete the question myself. Thank you.

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Let $\alpha_{0},\ldots,\alpha_{n}\in\mathbb{C}$ be given. Compute the spectrum of $T:\ell^{1}\rightarrow\ell^{1}$ with \begin{align} (Tx)_{i}=\sum_{j=0}^{n}\alpha_{j}x_{i+j}\text{ for all }i\geq 1. \end{align}

My Solution

I am only to find a bound on the spectrum. Let us look at the $\ell^{1}$ norm of the image of $T$. \begin{align} |Tx|_{1}&=\sum_{i=1}^{\infty}\bigg|\sum_{j=0}^{\infty}\alpha_{j}x_{i+j}\bigg|\\ &\leq \sum_{i=1}^{\infty}\sum_{j=0}^{\infty}|\alpha_{j}||x_{i+j}|\\ &= \sum_{i=1}^{\infty}(|\alpha_{0}||x_{i+0}|+|\alpha_{1}||x_{i+1}|+\ldots+|\alpha_{n}||x_{i+n}|)\\ &= |\alpha_{0}|\sum_{i=1}^{\infty}|x_{i}|+|\alpha_{1}|\sum_{i=1}^{\infty}|x_{i+1}|+\ldots+|\alpha_{n}|\sum_{i=1}^{\infty}|x_{i+n}|\\ &=|\alpha_{0}||x|_{1}+|\alpha_{1}||Lx|_{1}+|\alpha_{2}||L^{2}x|_{1}+\ldots+|\alpha_{n}||L^{n}x|_{1}\\ &\leq \sum_{j=0}^{n}|\alpha_{j}|, \end{align} where $L:\ell^{1}\rightarrow\ell^{1}$ is the left shift operator. By picking the sequence $(L^{n-1}x)_{n\geq 1}$ this shows that $\|T\|=\sum_{j=0}^{\infty}|\alpha_{j}|$. So $\text{spr}(T)\leq \sum_{j=0}^{\infty}|\alpha_{j}|$, where $\text{spr}(T)$ denotes the spectral radius of $T$.

Can someone give me some guidance on where to progress to from here? Please no answers, just hints.

Completion of the proof

Observe by expanding the series you can rewrite $Tx$ as the polynomial of some other operator. \begin{align} Tx &= (\sum_{j=0}^{n}\alpha_{j}x_{1+j},\sum_{j=0}^{n}\alpha_{j}x_{2+j},\sum_{j=0}^{n}\alpha_{j}x_{3+j},\ldots)\\ &= (\alpha_{0}x_{1}+\alpha_{1}x_{2}+\ldots+\alpha_{n}x_{n+1},\\ &\hspace{22.5mm} \alpha_{0}x_{2}+\alpha_{1}x_{3}+\ldots+\alpha_{n}x_{n+2},\\ &\hspace{37.5mm} \alpha_{0}x_{3}+\alpha_{1}x_{4}+\ldots+\alpha_{n}x_{n+3},\ldots)\\ &=(\alpha_{0}x_{1},\alpha_{0}x_{2},\ldots,\ldots) + (\alpha_{1}x_{2},\alpha_{1}x_{3},\ldots) + \cdots +(\alpha_{n}x_{n+1},\alpha_{n}x_{n+2},\ldots)\\ &=\alpha_{0}x + \alpha_{1}Lx + \alpha_{2}L^{2}x + \cdots +\alpha_{n}L^{n}x\\ &=(\alpha_{0}+\alpha_{1}L+\alpha_{2}L^{2}+\cdots+\alpha_{n}L^{n})x. \end{align}

Therefore the spectrum of $T$ is equal to the polynomial of the spectrum by the spectral mapping theorem. That is, \begin{align} \sigma(T)&=\sigma(\alpha_{0}+\alpha_{1}L+\alpha_{2}L^{2}+\cdots+\alpha_{n}L^{2})\\ &=\alpha_{0}\sigma(I)+\alpha_{1}\sigma(L)+\alpha_{2}\sigma(L^{2})+\cdots+\alpha_{n}\sigma(L^{n})\\ &= \alpha_{0}\sigma(I)+\alpha_{1}\sigma(L)+\alpha_{2}\sigma(L)^{2}+\cdots+\alpha_{n}\sigma(L)^{n}. \end{align}

Now, $\sigma(L)=\{\lambda\in\mathbb{C}||\lambda|\leq 1\}$, from a previous question I completed.Therefore, \begin{align} \sigma(T) = \{\alpha_{0}+\alpha_{1}\lambda +\alpha_{2}\lambda^{2}+\cdots+\alpha_{n}\lambda^{n}\in\mathbb{C}||\lambda|\leq 1\}. \end{align} We know $\text{spr}(T)\leq\sum_{j=0}^{n}|\alpha_{j}|$ and, \begin{align} \sup_{\lambda\in\sigma(T)}|\alpha_{0}+\alpha_{1}\lambda+\alpha_{2}\lambda^{2}+\cdots+\alpha_{n}\lambda^{n}|\leq|\alpha_{0}+\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}|\leq\sum_{j=0}^{n}|\alpha_{j}|, \end{align} hence we have found the entire spectrum.

Answer 1

I don't think there is an elementary way of doing this. The key observation is that $T=p(S)$, where $p$ is a polynomial and $S$ is the unilateral shift $$ Sx=(0,x_1,x_2,\ldots). $$ So the Spectral Mapping Theorem reduces the problem to finding the spectrum of $S$. This is still non-obvious, as far as I can tell, but one can consider the adjoint $S^*:\ell^\infty\to\ell^\infty$. The spectrum of $S^*$ is easy to calculate, as it is almost entirely made up of eigenvalues. Then using $\sigma(S)=\sigma(S^*)$ (another non-trivial fact for operators on Banach spaces as opposed to Hilbert spaces) we are done.