Compact operator without eigenvalues? 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?
 A: 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.
A: Sure, let $\mu_n = \frac{1}{|n|+ 1}$, then for each $n$ the vector $\sqrt{\mu_{n-1}}e_n\oplus \pm\sqrt{\mu_n}e_{n-1}$ is an eigenvector with eigenvalue $\pm\sqrt{\mu_n\mu_{n-1}}$.
