Consider the operator $M$ on $\ell^2(\mathbb{Z})$ defined by for $u\in \ell^2(\mathbb Z)$


$$Mu(n)=\frac{1}{\vert n \vert+1}u(n).$$ This is a compact operator!


Then, let $l$ be the left-shift and $r$ the right-shift on $\ell^2(\mathbb Z).$


We consider the compact operator on $\ell^2(\mathbb Z;\mathbb C^2)$ defined by


$$T:=\begin{pmatrix} 0 & l M \\ rM & 0 \end{pmatrix}$$


My question is: Even though $T$ is not normal, since $$T^\*T= \begin{pmatrix} MlrM & 0 \\ 0 & Mr lM \end{pmatrix}=M^2$$
whereas $$TT^\*= \begin{pmatrix} lM^2 r & 0 \\ 0 & rM^2 l \end{pmatrix}\neq M^2$$ does $T$ have eigenvalues?



Compact operator without eigenvalues?
Note that $$T^2 = \begin{pmatrix}lMrM&0\\0&rMlM\end{pmatrix},$$ and hence the eigenvectors of $T^2$ are $$v\_j = (e\_j, 0) , \qquad w\_j = (0, e\_j),$$ with corresponding eigenvalues $$\lambda\_j = \frac{1}{(1 + |j|) (1 + |j+1|)} \, , \qquad \mu\_j = \frac{1}{(1 + |j|) (1 + |j-1|)} \, ,$$ respectively. In particular, the eigenspaces of $T^2$ are four-dimensional: for $j \ge 0$, the eigenspace corresponding to $\lambda\_j = \mu\_{j+1} = \lambda\_{-j-1} = \mu\_{-j}$ is spanned by $v\_j$, $u\_{j+1}$, $v\_{-j-1}$ and $u\_{-j}$.


If $u$ is an eigenvector of $T$, then it is also an eigenvector of $T^2$, and hence it is a linear combination of $v\_j$, $u\_{j+1}$, $v\_{-j-1}$ and $u\_{-j}$ for some $j \ge 0$. By inspection, $$ T(a v\_j + b u\_{j+1} + c v\_{-j-1} + d u\_{-j}) = \frac{a u\_{j+1} + d v\_{-j-1}}{1 + |j|} + \frac{b v\_j + c u\_{-j}}{1 + |j + 1|} $$ corresponds to a simple block $4\times4$ matrix, and it is now an elementary exercise to find the eigenvectors.


