Linear operator is trace non-increasing if dual operator is sub-unital 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!
 A: 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":
$$
\operatorname{tr}\Big(L \begin{pmatrix}-1&0\\0&-1\end{pmatrix} \Big)=\operatorname{tr}\begin{pmatrix}-1&0\\0&0\end{pmatrix}=-1\ {\color{red}>}-2=\operatorname{tr}\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\,.
$$

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 $B\geq 0$.
