Eigenvalues of antiunitary operators

Question

I have sometimes come across the statement that antiunitary operators have no eigenvalues. For example, on page 34 in the book "Topological Insulators and Topological Superconductors" by Bernevig and Hughes, it is stated that

The preceding ( $T i T^{-1} = -i$ ) makes it clear that the time-reversal operator $T$ must be proportional to the operator of complex conjugation. Such operators are called antiunitary and, unlike unitary (sic.) operators, do not have eigenvalues.

I do not understand this statement. For example, consider the antiunitary operator $\sigma_x K$ where $K$ corresponds to complex conjugation and $\sigma_x$ is a Pauli matrix, then

\begin{equation} \sigma_x K \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix} = \pm \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix} \end{equation}

Naively, I would therefore conclude that $\left( 1, \pm 1 \right)^T$ is an "eigenstate" of $\sigma_x K$ with "eigenvalue" $\pm 1$. If we multiply this eigenstate by a phase $e^{i\phi}$, it remains an eigenstate but its "eigenvalue" changes by $e^{-2i\phi}$. Hence, it seems that one can have eigenstates of an antiunitary operator but their eigenvalue is not a single scalar. Instead the eigenvalue corresponds to a circle.

Edit

I have found this paper which deals with the subject, but seems to contradict the original statement: https://arxiv.org/abs/1507.06545

If we consider the time-reversal operator again, since for spinless particles $T^2=1$, there exist eigenstates of $T$ without unique eigenvalues. However, for spin 1/2 particles, $T^2 = -1$ and there exist no eigenstates (see the answer of CosmasZachos).

Answer 1

Your fine link has the answer for you in its section 2.2, illustrating that some antiunitary operators, like Fermi's spin flip, lack eigenvectors, as you may easily check.

But the counterexample you chose is of the $\vartheta ^2={\mathbb I}$ variety, and so $\vartheta$ does have the obvious eigenvectors: that's the point of Proposition 2.3 , corollary 2.4 !

Check your $$ \sigma_x K \sigma_x K ={\mathbb I}, $$ in sharp contrast to $$ i\sigma_y K i\sigma_y K =-{\mathbb I}. $$ The first has eigenvectors with $\vartheta^2$ having a positive semidefinite spectrum, but the second doesn't.