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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?

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  • $\begingroup$ What is $u$ in the definition of $M$? Where does $T$ act? $\endgroup$
    – Fedor Petrov
    Commented Jul 10, 2020 at 8:23
  • $\begingroup$ @FedorPetrov sorry, $u \in \ell^2(\mathbb Z)$ and $T$ acts on $\ell^2(\mathbb Z; \mathbb C^2).$ $\endgroup$
    – Landauer
    Commented Jul 10, 2020 at 8:34

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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.

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  • $\begingroup$ interesting, do you see any argument using abstract arguments that there have to be eigenvalues? It seems in this case, the simple structure of the square bailed us out-which I missed. But perhaps it is clear already from something that there has to be at least one? $\endgroup$
    – Landauer
    Commented Jul 10, 2020 at 10:36
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    $\begingroup$ @Martinique: Of course there are compact operators with no eigenvalues whatsoever. On the other hand, if any power of $T$ is normal, then the above argument works (the cyclic subspace generated by an eigenvector $v$ of some $T^n$, that is, $\operatorname{lin}\{v, Tv, T^2v, T^3v,\ldots\}$ is finite-dimensional, and so $T$ has an eigenvector within this subspace). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 10, 2020 at 11:23
  • $\begingroup$ There's a typo in the penultimate paragraph: 'If $u$ is an eigenvector of $T^2$, then it is also an eigenvector of $T^2$', while tautologically true, is certainly not what was meant there... :) I presume the first one is meant to be $T$ rather than $T^2$, since that implication is true and non-trivial... $\endgroup$
    – Steven Stadnicki
    Commented Jul 10, 2020 at 20:05
  • $\begingroup$ @StevenStadnicki: Hardly non-trivial, I'd say, but yes, this is what I meant. Thanks! $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 10, 2020 at 20:20
  • $\begingroup$ Small remark: In case that one is only interested in the existence of eigenvalues (as the OP seems to be, according to their comments), one can replace the second half of the argument by just noting that if $\lambda$ is an eigenvalue of $T^2$, than at least one of the complex roots of $\lambda$ is an eigenvalue of $T$ (which is true for every linear operator $T$). $\endgroup$
    – Jochen Glueck
    Commented Jul 10, 2020 at 21:51
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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 $\pm\sqrt{\mu_n\mu_{n-1}}$.

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  • $\begingroup$ Thank you also for your answer. Since Mateusz answered a few minutes earlier, I just accepted his answer, I hope this is okay. But let me also ask you, if you see any abstract argument that there have to be eigenvalues (so not by explicitly diagonalization) but by some a priori argument? $\endgroup$
    – Landauer
    Commented Jul 10, 2020 at 10:37
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    $\begingroup$ Well, under what assumptions? My construction works for your operator with any sequence in $c_0(\mathbb{Z})$ in place of $(\mu_n)$. In general every nonzero element of the spectrum of a compact operator is an eigenvalue, but there do exist compact operators with spectrum $\{0\}$ and no eigenvalues. $\endgroup$
    – Nik Weaver
    Commented Jul 10, 2020 at 11:10

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