Linear operator is trace non-increasing if dual operator is sub-unital

Question

I am dealing with the following problem and hope some of you can help me.

Consider two finite dimensional Hilbert spaces $A$ and $B$ and a linear map $L \in \mathcal{L}(A) \rightarrow \mathcal{L}(B)$. The dual map $L^*$ is defined via the Hilbert-Schmidt inner product, namely $\forall M \in \mathcal{L}(A) ~\forall N \in\mathcal{L}(B): ~Tr[N^* L(M)] = Tr[(L^*(N))^* M]$. Furthermore, we define

$A \leq B ~:\Leftrightarrow B-A$ is positive semi-definite

I want to show the following statement.

(i) $\forall M \in \mathcal{L}(A): ~ Tr[L(M)] \leq Tr[M]$ $\Leftrightarrow$ $L^*(I^B) \leq I^A$

So, I have to show two directions. I cannot relate the positivity to a condition under the trace (and following the definition of the dual, I only have statements under the trace).

e.g. for (i) $"\Rightarrow"$ I tried the following:

I started with setting $N=I^B$ and obtained from the definition of the dual

$Tr[I^B L(M)] = Tr[(L^*(I^B))^*M]$

$\Leftrightarrow Tr[M] \geq Tr[L(M)] = Tr[(L^*(I^B))^*M]$

This has to hold for arbitrary $M \in \mathcal{L}(A)$, so

$\Leftrightarrow \forall M \in \mathcal{L}(A): ~ Tr[(I^A - L^*(I^B))M] \geq 0 $.

Let $\{ v_1,...,v_n\}$ be the set of eigenvectors of $(I^A - L^*(I^B))$, where $n = dim(A)$. Now I can choose $M$ to be the identity matrix and take the trace with respect to the eigenbasis $\{ v_1,...,v_n\}$, then I find that all eigenvalues of $(I^A - L^*(I^B))$ are positive, hence we have shown this direction.

Unfortunately, I do not know anything about the geometric multiplicity of the eigenvalues of $(I^A - L^*(I^B))$, hence we do not even know whether we find an eigenbasis. So, in general, my argument seems to be wrong.

Does anyone know how to prove this statement? I am grateful for any help/hint!

Answer 1

Your statement (i) as it stands is wrong. To see this consider the map $$ L:\mathbb C^{2\times 2}\to \mathbb C^{2\times 2}\quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\mapsto \begin{pmatrix}1&0\\0&0\end{pmatrix}^*\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\0&0\end{pmatrix} $$ which is obviously

• linear
• completely positive (due to Kraus form)
• self-dual, that is, $L=L^*$ (if $L(A)=K^*AK$ then $L^*(B)=KBK^*$, but here the Kraus operator is self-adjoint: $K=K^*$).

In particular, $L^*$ is sub-unital because $L$ is (self-duality). However---despite $L^*$ being sub-unital---$L$ violates your definition of "trace non-increasing": $$ \operatorname{tr}\Big(L \begin{pmatrix}-1&0\\0&-1\end{pmatrix} \Big)=\operatorname{tr}\begin{pmatrix}-1&0\\0&0\end{pmatrix}=-1\ {\color{red}>}-2=\operatorname{tr}\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\,. $$

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What you really want to show is the following statement: $$ \forall_{M\in\mathcal L(A){\bf, M\geq 0}}\ \operatorname{tr}(L(M))\leq\operatorname{tr}(M)\ \Leftrightarrow\ L^*(I^B)\leq I^A\,,\tag*{(1)} $$ that is, one only checks the trace condition for $M$ positive semi-definite. Then the proof of (1) becomes rather simple: \begin{align*} \forall_{M\geq 0}\ \operatorname{tr}(L(M))\leq\operatorname{tr}(M)&\ \Leftrightarrow\ \forall_{M\geq 0}\ \operatorname{tr}(ML^*(I^B))\leq\operatorname{tr}(M)\\ &\ \Leftrightarrow\ \forall_{M\geq 0}\ 0\leq \operatorname{tr}\big(M\big(I^A-L^*(I^B)\big)\big)\\ &\ \Leftrightarrow\ 0\leq I^A-L^*(I^B)\ \Leftrightarrow\ L^*(I^B)\leq I^A \end{align*} The only non-trivial (but not too difficult, either) step here is to show that a matrix $A$ is positive semi-definite if and only if $\operatorname{tr}(AB)\geq 0$ for all $B\geq 0$.