Let $A,B\in \mathbb{R}^{n\times n}$. And let $vec$ be a map : $\mathbb{R}^{n\times n}\to \mathbb{R}^{n^2\times1}$: $vec(e_a e_b^T)=e_a\otimes e_b$, here $e_a$ denotes the vector with with an entry $1$ on index $a$ and other entries $0$. Intuitively the operation $vec$ puts every row of a matrix into a new vector. Partial trace is defined as $tr_B(A\otimes B)=tr(B)A$. Then how to prove $tr_B(vec(A)vec(B)^T)=AB^T$? Here $\otimes$ denotes the Kronecker product.
Attempts: I try to decomposite $vec(A)vec(B)^T$ into the form $\sum_{ij} X_i\otimes Y_j$ so $tr_B(vec(A)vec(B)^T)=\sum_{ij}(X_i tr(Y_j))$ , but how to achieve that?
I have one new idea: $vec(A)=\sum_{ij}a_{ij} e_i\otimes e_j$ and $vec(B)=\sum_{kl}b_{kl}e_k\otimes e_l$, I will have a try along this way.
Best Answer
I know how to prove now. $vec(A)=\sum_{ij}a_{ij}e_i\otimes e_j$, $vec(B)=\sum_{kl}b_{kl}e_k\otimes e_l$ \begin{align*} tr_B(vec(A)vec(B)^T)&=tr_B(\sum_{ijkl}a_{ij}b_{kl}(e_i\otimes e_j)(e_k^T\otimes e_l^T))\\ &=tr_B(\sum_{ijkl}a_{ij}b_{kl}(e_i e_k^T)\otimes(e_je_l^T))\\ &=\sum_{ijk}a_{ij}b_{kj}e_ie_k^T=\sum_{ijk}e_i(e_i^TAe_j)(e_j^TB^Te_k)e_k^T=AB^T \end{align*}