Partial trace and linear operators Trying to solve the following:
Consider a linear map as follows 
$$vec: L(X,Y)\rightarrow Y \otimes X $$
$$ vec: E_{b,a} \mapsto e_b \otimes e_{a}$$ which can be looked as a change of a basis map. Where $E_{b,a}$ is usual basis for $L(X,Y)$ given by $$E_{b,a}(d,c)=1\ \text{iff } a=c\ \&\ b=d\ \\  \hspace{1cm}0\ otherwise.$$
and $e_{b}$ and $e_{a}$ are usual basis elements for $Y$ and $X$ respectively. 
In bra-ket notation this can be written down as $vec \left( \vert  e_{b}\rangle\langle e_{a}\vert \right) \mapsto \vert e_{b}\rangle \vert e_{a}\rangle.$ Let $A,B \in L(X,Y)$ then
$$Tr_Y ( vec ( A ) vec ( B )^{*} ) = ( B^* A )^T $$.
Here is my failed attempt:
$A=\Sigma A(y,x)E_{y,x}$ and $vec(A)=\Sigma A(y,x)e_{y} \otimes e_{x}$
$B=\Sigma B(y,x)E_{y,x}$ and $vec(B)^{*}=\Sigma\overline{B(y,x)}e_{y}^{*}\otimes e_{x}^{*}$.
Hence $Tr_Y ( vec ( A ) vec ( B )^{*})(x,x_2) = \Sigma_{y} A(y,x) \overline{B(y,x_{2})}$
 and 
$B^{*}(x,y)=\Sigma \overline{B(x,y)}E_{x,y}$ giving 
$B^*A(x_{2},x)=\Sigma \overline{B(x_{2},y)} A(y,x)$ which gives
$(B^*A)^T(x,x_2)=\Sigma \overline{B(x_{2},y)} A(y,x)$.
As you can see arguments of B are reversed in my attempt. Where did I go wrong?
 A: With
$$
A = \sum_{a}\sum_{b}A(a,b)E_{a,b} \qquad \text{and}\qquad B = \sum_{c}\sum_{d}B(c,d)E_{c,d}
$$
one has 
$$
\operatorname{vec}(A) = \sum_{a}\sum_b A(a,b) \, e_a\otimes e_b
\qquad \text{and}\qquad B = \sum_{c}\sum_{d}B(c,d)\,e_c\otimes e_d
$$
such that
$$
\operatorname{vec}(A)\operatorname{vec}(B)^* = \sum_{a,b,c,d}A(a,b)\overline{B(c,d)}\, E_{a,c} \otimes E_{b,d}.\tag{$\ast$} \label{1}
$$
For any pair of indices $a$ and $c$, note that
$$
\operatorname{Tr}(E_{a,c}) = \left\{\begin{array}{ll}
 1 & \text{if }a=c,\\
 0 & \text{otherwise}.
\end{array}\right.
$$
Taking the partial trace of $\operatorname{vec}(A)\operatorname{vec}(B)^*$ in \eqref{1} therefore yields
$$
\operatorname{Tr}_\mathcal{Y}(\operatorname{vec}(A)\operatorname{vec}(B)^*) = \sum_{b,d}\sum_{a} A(a,b)\overline{B(a,d)}\, E_{b,d}.
$$
Note that $(B^*A)^{\scriptscriptstyle \mathsf{T}} = A^{\scriptscriptstyle \mathsf{T}} \overline{B}$. It remains to show that $(A^{\scriptscriptstyle \mathsf{T}} \overline{B})(b,d) = \sum_{a}A(a,b)\overline{B(a,d)}$.
