Spectrum of operator $T((x_n)_{n\in\mathbb{Z}})=\left(\frac{1}{n^2+1}(x_n-x_{-n})\right)_{n\in\mathbb{Z}}$

Question

The eigenvalues should satisfy: $$T(x_n)=\lambda x_n$$ $$\frac{1}{n^2+1}(x_n-x_{-n})=\lambda x_n$$ $$\left[(n^2+1)\lambda+1\right]x_n=x_{-n}$$ I suppose that this should mean that $$\forall\lambda\in\mathbb{C}\ \exists(x_n)\neq0:T((x_n))=\lambda(x_n)$$ but my teachers solution says the point spectrum is $\{0\}\cup\{\frac{2}{n^2+1};\ n\in\mathbb N \}$.

Where did I go wrong?

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$T:l^1(\mathbb{Z})\rightarrow l^1(\mathbb{Z})$

Answer 1

Hint: if $[1-\lambda(n^2+1)]x_n=x_{-n}$ for all $n$, then $[1-\lambda(n^2+1)]x_{-n}=x_{n}$ for all $n$ as well.

Answer 2

The sequence $x=\{ \cdots,0,0,-1,0,1,0,0,\cdots\}$ where the middle $0$ is $x_0$ is mapped to $\{ \cdots,0,0,\frac{-2}{1^2+1^2},0,\frac{2}{1^2+1^2},0,0,\cdots\}$ under the transformation $T$. That is $Tx=\frac{2}{1^2+1^2}x$. In this way you can see that $\frac{2}{1+n^2}$ is an eigenvalue for $n=1,2,3,\cdots$. And $0$ is an eigenvalue also because $Ty=0$ where $y=\{\cdots,0,0,1,0,1,0,0,\cdots\}$.

Let $x_n$ denote the sequence with all $0$'s except for a $-1$ in the $-n$ position, and $1$ in the $+n$ position. And let $y_n$ denote the sequence with all $0$'s except for a $1$ in the $-n$ position and a $1$ in the $n$ position. Then $$ Tx_n = \frac{2}{n^2+1} x_n,\;\;\; Ty_n = 0. $$ You can't have any other point spectrum because the set of all $x_n$ and $y_n$ together form a complete orthogonal basis of $\ell^2$. In fact, if $e_n$ is the standard basis element with $1$ in the $n$-th place and $0$'s elsewhere, then $e_n = \frac{1}{2}(x_n+y_n)$ if $n > 0$ and $e_n = \frac{1}{2}(y_{-n}-x_{-n})$ if $n < 0$. You can use this fact to show that there cannot be any other eigenvalues of $T$.