# Are peripheral eigenvalues of a completely positive map always semisimple?

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.

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

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.