Consider a linear map as follows
$$vec: L(X,Y): Y \otimes X $$ $$ vec: E_{b,a} \mapsto e_b \otimes e_{a}$$ 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 $vec \left( \vert e_{b}\rangle\langle e_{a}\vert \right) \mapsto \vert e_{b}\rangle \vert e_{a}\rangle.$
trying to prove following :
- Let $\ X_{1},Y_{1},X_{2},Y_{2}\ $ be complex Euclidean spaces. Then for every choice of operators A ∈ L $(X_1 , Y_1 ) ,\ B ∈ L (X_2 , Y_2 ) ,\ and\ X ∈ L (X_ 2 , X_1 ) $,
it holds that ( A ⊗ B ) vec ( X ) = vec ( $AXB^T$ ).
- $u\in X$ and $v \in Y$ we have $$vec(uv^{*})= u \otimes \overline{v}.$$
Any hints are welcome.
One statement follows easily from the other, so I'll prove statement 2 and give you a hint to prove statement 1.
The vectors $u$ and $v$ can be expanded in the standard bases of their respective spaces as $$ u=\sum_{a}u(a)e_a \qquad\text{and}\qquad v=\sum_{b}v(b)e_b . $$ Note that $e_ae_b^* = E_{a,b}$. By linearity of $\mathrm{vec}$, one has that \begin{align*} \mathrm{vec}(uv^*) &= \sum_{a}\sum_{b}u(a)\overline{v(b)} \operatorname{vec}(e_ae_b^*) \\ & = \sum_{a}\sum_{b} u(a)\overline{v(b)}\, e_a\otimes e_b\\ & = \Bigl(\sum_a u(a)e_a \Bigr) \otimes \Bigl(\sum_{b}\overline{v(b)}e_b \Bigr)= u\otimes \overline{v}, \end{align*} as desired.
To prove statement 1, expand the operator $X$ as $$ X = \sum_{a,b} X(a,b) \, E_{a,b}. $$ One then has that $$ A\otimes B \operatorname{vec}(X) =(A\otimes B)\sum_{a,b}X(a,b) e_a\otimes e_b= \sum_{a,b}X(a,b) (Ae_a)\otimes (Be_b). $$ Use statement 2, and the fact that $(\overline{Be_b})^* = e_b^*B^{\scriptscriptstyle\mathsf{T}}$, to complete the proof.