# Do completely positive maps have a leading eigenoperator with nonvanishing trace?

A completely positive map $$\mathbb{W}$$ is a map from $$\mathbb{C}^{n\times n} \to \mathbb{C}^{n\times n}$$ that can be written in terms of $$n\times n$$ matrices $$K$$ as

$$\mathbb{W}(\rho) = \sum_{i=1}^N K_i \rho K^\dagger_i$$

Note that I do not require $$\sum_i K_i^\dagger K_i = I$$, so this is generically not a quantum channel, as it is generically not trace preserving (and hence generically doesn't take density matrices to density matrices).

Consider the set of eigenvalues of $$\mathbb{W}$$, and consider the eigenvalue $$\lambda_m$$ with the maximum absolute value.

Am I guaranteed that there exists at least one $$\rho_m$$ such that $$\mathbb{W}(\rho_m) = \lambda_m \rho_m$$ and $$Tr[\rho_m] \neq 0$$ are both satisfied?

That is, can I always find a not-traceless matrix in the leading-eigenvalue eigenspace of $$\mathbb{W}$$?

I actually suspect I might be able to ask for even more properties, like at least one positive semi-definite matrix in the leading-eigenvalue eigenspace of $$\mathbb{W}(\rho)$$. However, for my purposes, I'm most interested in the weakened statement above of a not-traceless matrix.

• Those are called completely positive maps. (I sometimes also call them quantum channels, though I think the mainstream use of the latter includes trace preserving.) Oct 24, 2023 at 8:37
• @NorbertSchuch Thanks for the edits! Oct 24, 2023 at 19:01

There, Theorem 6.5 states that any positive map (and thus, in particular, any completely positive map) with spectral radius $$\varrho$$ has $$\varrho$$ as an eigenvalue, and there is a positive semi-definite $$X$$ which is an eigenvector with that eigenvalue.
One intuition of why this is the case might be taken from an iterated application of the map $$X\mapsto \mathbb W(X)/|\lambda_m|$$, which will converge to an eigenvector with eigenvalue $$\lambda_m$$ for any initial $$X$$. On the other hand, for a positive $$X$$, this map will only ever return positive semi-definite matrices. What remains open in this argument is to show that there is at least one $$X\ge0$$ which has non-zero overlap with a fixed point.
• Convergence to an eigenvector with maximal eigenvalue will happen only for initial $X$ that have overlap with an eigenvector with maximal eigenvalue. If the identity $I$ is such a vector, we get using positivity that the maximal-eigenvalue eigenvector it has overlap with is itself positive semi-definite. cont'd Oct 24, 2023 at 18:32
• If the identity $I$ is not such a vector, consider a vector $X$ that does have overlap with an eigenvector of maximal eigenvalue, and note $X+a I$ will converge to that eigenvector under iterated application of the scaled map for any $a$. In particular, we can take $a$ large and positive, and hence the maximal-eigenvalue eigenvector must itself be positive semi-definite. Oct 24, 2023 at 18:32