Are peripheral eigenvalues of a completely positive map always semisimple?

Question

It is known that all peripheral eigenvalues (i.e. all eigenvalues $\lambda\in\mathbb C$ such that $|\lambda|$ equals the spectral radius) of positive trace-preserving or positive unital maps are always semisimple (=no non-trivial Jordan blocks), cf. Proposition 6.2 in the lecture notes of Michael Wolf.

In some sense this is a generalization of the fact that the peripheral eigenvalues of any stochastic matrix are always semisimple, as well as the Perron-Frobenius theorem which states that for positive matrices the leading eigenvalue is always simple. Now, motivated by the latter result the following question arises:

Given $\Phi\in\mathcal L(\mathbb C^{n\times n})$ completely positive (but not necessarily trace preserving) are the peripheral eigenvalues of $\Phi$ always semisimple? Or if complete positivity is not enough, maybe strict positivity—as an analogue of the requirement for Perron-Frobenius—guarantees such a result, maybe even just for the leading eigenvalue?

In other words the question is whether trace-preservation, resp. unitality was a necessary assumption in the original result on peripheral eigenvalues of positive, trace-preserving maps, or whether (complete, strict, or just usual) positivity is all one needs. As a side note this question is of course related to the fact that not every quantum channel is diagonalizable: in the diagonalizable case all eigenvalues are semisimple so any possible counterexample to the above question has to be a non-diagonalizable map.

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(This is a Q&A style question meant as a contribution to the list of counterexamples in quantum information)

Answer 1

Consider $$ K:=\begin{pmatrix}1&1\\0&1\end{pmatrix} $$ as well as $\Phi:=K(\cdot)K^\dagger$. This map is completely positive (because $\Phi$ is in Kraus form) and even strictly positive because $$ \Phi({\bf1})=KK^\dagger=\begin{pmatrix}2&1\\1&1\end{pmatrix}>0\,. $$ However, the representation matrix $\widehat\Phi$ of $\Phi$—i.e. the unique $4\times 4$-matrix which satisfies ${\rm vec}(\Phi(X))=\widehat\Phi{\rm vec}(X)$ for all $X\in\mathbb C^{2\times 2}$ with ${\rm vec}$ the usual vectorization operation—reads $$ \widehat\Phi=\overline{K}\otimes K=\begin{pmatrix}1&1&1&1\\0&1&0&1\\0&0&1&1\\0&0&0&1\end{pmatrix}=SJS^{-1} $$ where its Jordan decomposition is induced via $$ S=\begin{pmatrix} -\frac12&2&1&0\\ \frac12&0&1&0\\ -\frac12&0&1&0\\ 0&0&0&1 \end{pmatrix}\quad\text{ and }\quad J=\begin{pmatrix} 1&0&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1 \end{pmatrix}. $$ Thus the leading (and in fact only) eigenvalue of $\Phi$ is $1$, but that eigenvalue is not semisimple because its geometric multiplicity is $2$ (two-dimensional eigenspace) is strictly smaller then its algebraic multiplicity $4$.

Moreover, this example shows that a completely positive map can have spectrum $\{1\}$ without being the identity map—a conclusion which one can only draw if the map were additionally trace-preserving or unital.