Positive and trace-preserving transformations with a common fixed point of full rank

The following problem which has been on my mind for a while now arises from the realm of quantum information involving quantum channels with a common fixed point of full rank, as well as majorization theory, but can really be boiled down to a problem in linear algebra.

One considers maps acting on complex $$n\times n$$ square matrices which are linear, trace-preserving and positive (more precisely "positivity-preserving", i.e. if $$A\in\mathbb C^{n\times n}$$ is positive semi-definite, then so is its image under the map in question). Let us denote the convex set of all such maps by $$P(n)\subset\mathcal L(\mathbb C^{n\times n})$$ where the latter is the vector space of all linear maps/operators on $$\mathbb C^{n\times n}$$.

Conjecture. Let $$A,B,X\in\mathbb C^{n\times n}$$ with $$A,B$$ hermitian and $$X$$ positive definite. If there exist $$T_1,T_2\in P(n)$$ with $$T_1(X)=T_2(X)=X,\quad T_1(A)=B\quad\text{and}\quad T_2(B)=A,$$ then there exists unitary $$U\in\mathbb C^{n\times n}$$ such that $$[U,X]=0$$ and $$U^\dagger AU=B$$.

The assumption of a common fixed point is essential. To see this, consider arbitrary $$\rho_1,\rho_2\in\mathbb C^{n\times n}$$ positive semi-definite and of trace $$1$$. The maps $$T_i:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$$, $$X\mapsto \rho_{2-i} \operatorname{tr}(X)$$ are in $$P(n)$$ and satisfy $$T_1(\rho_1)=\rho_2$$, $$T_2(\rho_2)=\rho_1$$, but in general $$\rho_1,\rho_2$$ (of course) are not unitarily equivalent. So the common fixed point seems to rule out such "projective" maps from $$P(n)$$ which only have one-dimensional range.

Note that maps of the above form for example come up in majorization theory; for two hermitian matrices $$A,B$$ one can show that $$A$$ majorizes $$B$$ (comparison of the vectors of eigenvalues) if and only if there exists $$T\in P(n)$$ with $$T(\operatorname{id})=\operatorname{id}$$ and $$T(A)=B$$, cf. e.g. Theorem 7.1 in the linked manuscript. In such a framework (that is, $$X=\operatorname{id}$$ and $$A,B$$ hermitian), one quickly sees that the above conjecture in fact holds as then $$\sigma(A)=\sigma(B)$$. The arising difficulty for $$X\neq \operatorname{id}$$ in some sense comes from the loss os symmetry: a map $$T\in\mathcal L(\mathbb C^{n\times n})$$ is trace-preserving if and only if its dual $$T^*$$ (w.r.t. the Hilbert-Schmidt inner product) is unital/identity-preserving. Thus one may reformulate the conjecture as follows.

Conjecture (v2). Let $$A,B,X\in\mathbb C^{n\times n}$$ with $$A,B$$ hermitian and $$X$$ positive definite. If there exist positive, linear maps $$T_1,T_2:\mathbb C^{n\times n}\to \mathbb C^{n\times n}$$ such that $$T_1(X)=T_2(X)=X,\quad T_1^*(\operatorname{id})=T_2^*(\operatorname{id})=\operatorname{id},\quad T_1(A)=B\quad\text{and}\quad T_2(B)=A,$$ then there exists unitary $$U\in\mathbb C^{n\times n}$$ such that $$[U,X]=0$$ and $$U^\dagger AU=B$$.

One readily verifies that there is a one-to-one connection between unital maps from $$P(n)$$ (as used in majorization theory) and positive, linear maps satisfying $$T(Y)=Y$$ and $$T^*(Y^{-1})=Y^{-1}$$ for some $$Y>0$$ via the transformation $$\Psi:\mathcal L(\mathbb C^{n\times n})\to P(n)\qquad T\mapsto Y^{-1/2}T(Y^{1/2}(\cdot)Y^{1/2})Y^{-1/2}$$ However, I do not see if and how this could be connected to the kind of maps from Conjecture (v2): $$T(\operatorname{id})=T^*(\operatorname{id})=\operatorname{id} \overset\checkmark\leftrightarrow T(X^{1/2})=X^{1/2},T^*(X^{-1/2})=X^{-1/2} \overset?\leftrightarrow T(X)=X,T^*(\operatorname{id})=\operatorname{id}$$

As a final remark to this post: the above conjecture, if true, seems to be closely related to Wigner's theorem for state-automorphisms$${}^1$$ or more generally linear isometries between matrix algebras. I am also aware of results on quantum channels and full-rank fixed points, cf. Chapter 6.4 in this script on quantum channels & operations by M. Wolf as well as this paper. However, so far I was not yet able to solve this puzzle. Thanks in advance for any answer or comment!

P.S. I hope the tags I chose for this question are fitting - but I am of course welcoming any comment addressing whether the tags are suitable or not, or if they can be extended in a meaningful way!

Footnote $${}^1$$: Another idea could be, starting from $$T_1,T_2$$, construct one map in $$P(n)$$ which maps $$A$$ to $$B$$, $$B$$ to $$A$$, $$X$$ to $$X$$ and acts bijectively on $$\operatorname{span}\lbrace A,B,X\rbrace$$ or perhaps on the convex, compact set of all density matrices (positive semi-def. & trace $$1$$). However, I do not see yet how one would go on about such a construction.

Edit: Due to the comment of Josiah Park, let me share my efforts on the case $$n=2$$. Let $$A,B,X\in\mathbb C^{2\times 2}$$ such that $$A,B$$ hermitian and $$X>0$$. Note that $$T(A)=B$$ (as well as $$S(B)=A$$) trivially implies $$\operatorname{tr}(A)=\operatorname{tr}(B)=:c\in\mathbb R$$ so using linearity of $$T,S$$ and the common fixed point $$X$$, we may go to the trace-less hyperplane of hermitian matrices via $$T\big(\underbrace{A-\tfrac{c}{\operatorname{tr}(X)}X}_{=:\hat A}\big)=\underbrace{B-\tfrac{c}{\operatorname{tr}(X)}X}_{=:\hat B}\qquad S\big(B-\tfrac{c}{\operatorname{tr}(X)}X\big)=A-\tfrac{c}{\operatorname{tr}(X)}X\,.$$ As $$\hat A,\hat B$$ are hermitian and trace-less, there exist $$a,b\geq 0$$ and unitaries $$V_a,V_b$$ such that $$\hat A=V_a\operatorname{diag}(a,-a)V_a^\dagger$$ and $$\hat B=V_b\operatorname{diag}(b,-b)V_b^\dagger$$. Due to Theorem 3.1 in this paper by Perez-Garcia, Wolf, Petz and Ruskai $$T,S$$ for $$n=2$$ are $$p$$-norm contractive on the trace-less hermitian hyperplane for all $$p\in[1,\infty]$$. This yields $$\|\hat B\|_\infty= \|T(\hat A)\|_\infty\leq \|T\|_{\infty\to\infty}\|\hat A\|_\infty =\|\hat A\|_\infty = \|S(\hat B)\|_\infty\leq \|S\|_{\infty\to\infty}\|\hat B\|_\infty =\|\hat B\|_\infty$$ so $$a=\|\hat A\|_\infty=\|\hat B\|_\infty=b$$ and $$\hat A,\hat B$$ are unitarily equivalent. This translates to the existence of unitary $$U$$ with $$U^\dagger AU-\tfrac{c}{\operatorname{tr}(X)}U^\dagger XU=U^\dagger \big(A-\tfrac{c}{\operatorname{tr}(X)}X\big) U=U^\dagger \hat A U=\hat B=B-\tfrac{c}{\operatorname{tr}(X)}X$$ so $$U^\dagger AU=B-\tfrac{c}{\operatorname{tr}(X)}U^\dagger [U,X]\,.$$ If $$X\propto\operatorname{id}$$ we are done. However, if $$X\not\propto\operatorname{id}$$ then one yet has to show that the non-empty set $$\lbrace U\in\mathcal SU(2)\,|\, U^\dagger AU=B-\tfrac{c}{\operatorname{tr}(X)}U^\dagger [U,X]\,\rbrace$$ also is non-empty under the additional constraint $$[U,X]=0$$ - or maybe the original $$U$$ for some reason already satisfies $$[U,X]=0$$? This I sadly do not see at the moment - on the other hand, the full-rank condition of $$X$$ did not yet come into play it seems as we only used $$X$$ positive semi-definite & $$X\neq0$$.

Anyways, this approach is very explicit as in low dimensions such calculations are quite simple and contractivity of positive, trace-preserving maps in the $$\infty$$-/operator norm holds for $$n=2,3$$ (but fails for $$n>3$$; however, all counter-examples to this I know of do not have any matrix of full rank in their range). Also given how I used it, I am not sure how useful the operator norm for larger $$n$$ is regardless.

• Has the conjecture been verified for small $n$? – Josiah Park Dec 20 '18 at 7:15
• Good point @JosiahPark! I in fact have not been able to prove this even for $n=2$, but I edited my post to share some of my efforts in this simpler case. – Frederik vom Ende Dec 21 '18 at 8:40

No, even on $$D_3:={\mathbb C} \oplus {\mathbb C} \oplus {\mathbb C}$$ (the diagonal of $$M_3({\mathbb C})$$. Let $$e_0,e_1,e_2$$ denote the standard basis for $$D_3$$ and put $$T\colon e_0\mapsto \frac{1}{2}(e_1 + e_2),\, e_1\mapsto e_0,\,e_2\mapsto e_0$$ on $$D_3$$ (or on $$M_3({\mathbb C})$$ by ignoring the off-diagonal entries). Then, $$T$$ is a trace-preserving positive map which fixes $$\mathrm{diag}(2,1,1)$$ and flips $$\mathrm{diag}(2,0,0)$$ and $$\mathrm{diag}(0,1,1)$$.
• Splendid example, which (fun fact) even disproves an old result about vector-$d$-majorization when slightly modifying it: choose $d=(3,2,1)$ and $T(x,y,z):=(y+z,2x/3,x/3)$ so $Td=d$, $T$ is trace-preserving (i.e. preserves the sum of vector entries) but $T$ flips $(3,0,0)$ and $(0,2,1)$ which shows that $d$-majorization contrary to popular belief is not a partial ordering even if the entries of the weight vector $d$ are pairwise different. Anyways great work, thank you very much! – Frederik vom Ende Feb 7 at 8:32