# Can the spectral radius of a completely positive map exceed the spectral radius of its transition matrix?

Recalling the spectral radius $$r(T):=\max_{\lambda\in\sigma(T)}|\lambda|$$ of a linear map $$T$$ (where $$\sigma(T)$$ refers to the spectrum of $$T$$), it is known that every quantum channel $$\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$$ satisfies $$r(\Phi)=1$$. One can see this by combining the following two facts:

• Every channel is trace-norm contractive (arXiv) so $$\Phi(X)=\lambda X$$ for any $$X\neq 0$$ implies $$|\lambda|=\frac{\|\Phi(X)\|_1}{\|X\|_1}\leq\frac{\|X\|_1}{\|X\|_1}=1$$
• Every channel has a fixed point $$\Phi(Z)=Z$$ (cf. also Theorem 4.24 in Watrous' book) so $$r(\Phi)\geq 1$$; actually, one has the even stronger statement that all peripheral eigenvalues of positive trace-preserving (or positive unital) maps are always semisimple—cf. Prop. 6.2 in the lecture notes of Michael Wolf—but that's beyond our needs.

On the other hand given any orthonormal basis $$\{|g_j\rangle\}_j$$ of $$\mathbb C^n$$ the corresponding transition matrix $$T:=(\langle g_j|\Phi(|g_k\rangle\langle g_k|)|g_j\rangle)_{j,k=1}^n$$ of any channel $$\Phi$$ is a (left-)stochastic matrix meaning $$r(T)=1$$ by, e.g., the Gershgorin circle theorem (or the simple-to-verify fact that any stochastic matrix is $$1$$-norm contractive) together with the Perron-Frobenius theorem. Therefore $$r(\Phi)=1=r(T)$$ for all channels $$\Phi$$ meaning the "classical action" $$T$$ of any channel $$\Phi$$ cannot "exceed the eigenvalue power" of the original channel.

Another way to formulate this problem is to define the full dephasing channel $$D$$ with respect to the orthonormal basis $$\{|g_j\rangle\}_j$$ (i.e. all off-diagonal elements in this basis are set to zero) and to see that $$T$$ is essentially the same as $$D\circ\Phi\circ D$$; in any case their spectral radius $$r(\Phi)=r(D\circ\Phi\circ D)$$ certainly coincides. For me this begs the question if there is a singular fundamental channel property which is responsible for this spectral equality. In other words this for me motivates the following:

Question. Given any completely positive map $$\Phi:\mathbb C^{n\times n}\to\mathbb C^{n\times n}$$ does it hold that $$r(T)\leq r(\Phi)$$ regardless of the orthonormal basis used for the transition matrix $$T$$? Equivalently, is it always true that $$r(D\circ\Phi\circ D)\leq r(\Phi)$$?

As for some more motivation: dephasing $$D$$ is a unital channel so its natural representation $$\widehat D\in\mathbb C^{n^2\times n^2}$$ has largest singular value $$1$$. In this sense it is never "expansive" (actually: in any norm) so to me it makes sense to assume that applying $$D$$ to any map cannot enlarge any eigenvalues.

This is furthered by the fact that if $$\Phi$$ is self-adjoint—i.e. $${\rm tr}(A^\dagger\Phi(B))={\rm tr}((\Phi(A)^\dagger B)$$ for all $$A,B$$ or, equivalently, $$\widehat\Phi^\dagger=\widehat\Phi$$—then the above inequality is true. This follows, e.g., from the Poincaré separation theorem (also known as Cauchy's interlacing theorem, cf. Corollary III.1.5 in Bhatia's book "Matrix Analysis") which implies that for Hermitian matrices—such as $$\widehat\Phi$$ here—the largest eigenvalue of any compression of $$\widehat\Phi$$ (e.g., $$T$$) is upper bounded by the largest eigenvalue of $$\widehat\Phi$$. Because the spectral radius of $$\Phi$$ and $$T$$ are both attained on a positive eigenvalue ("Perron-Frobenius", due to positivity) this shows $$r(D\circ\Phi\circ D)=r(T)=\lambda_{\rm max}(T)\leq\lambda_{\rm max}(\widehat\Phi)=\lambda_{\rm max}(\Phi)=r(\Phi)\,,$$ as desired. This does not carry over to the non-Hermitian case in any way meaning it remains open for now.

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

While the inequality in quesiton does hold for all channels and all self-adjoint positive maps (as shown in the above question) perhaps surprisingly it fails for general completely positive maps. For this consider the qubit map $$\Phi(X):=KXK^*$$ with $$K=\begin{pmatrix}i&1\\1&0\end{pmatrix} \quad\Rightarrow\quad \widehat\Phi=\overline{K}\otimes K=\begin{pmatrix} 1&-i&i&1\\ -i&0&1&0\\ i&1&0&0\\ 1&0&0&0 \end{pmatrix}\,.$$ Then all elements of $$\sigma(\Phi)=\sigma(\widehat\Phi)=\{1,1,\frac{-1+i \sqrt{3}}{2},\frac{-1-i \sqrt{3}}{2} \}$$ have absolute value $$1$$ so $$r(\Phi)=1$$. However, the transition matrix $$T:=(\langle j|\Phi(|k\rangle\langle k|)|j\rangle)_{j,k=1}^2=\begin{pmatrix}1&1\\1&0\end{pmatrix}$$ has eigenvalues $$\sqrt{\frac12(3\pm\sqrt5)}$$ so $$r(T)=\sqrt{\frac12(3+\sqrt5)}=1.61803\ldots>1=r(\Phi)\,.$$ This is also a counterexample to Cauchy's interlacing theorem for non-Hermitian matrices.