How can I show that $(a^\top \otimes b)(b^\top \otimes a) = (a \otimes b)(b \otimes a)^\top$?

Question

Let symbol $\otimes$ denotes the Kronecker product, $a \in \mathbb R^n$ and $b \in \mathbb R^m$.

How can I show that $(a^\top \otimes bb^\top \otimes a) = (ba^\top \otimes ab^\top)$ ?

My final goal is to show that the commutation matrix $K^{m,n}$ can be computed in various ways, including: \begin{gather} \sum_{i=1}^n \sum_{j=1}^m e_{n,i} e_{m,j}^\top \otimes e_{m,j} e_{n,i}^\top \\ \sum_{i=1}^n e_{n,i} \otimes I_m \otimes e_{n,i}^\top \\ \sum_{j=1}^m e_{m,j}^\top \otimes I_n \otimes e_{m,j} \end{gather} The question I am asking is the only step where I'm a little stuck. And I have the feeling it's really easy.

Answer 1

I have it. As Expected it's really incredibly easy.

For vectors $a \in \mathbb R^n$ and $b \in \mathbb R^m$, $a^\top \otimes b = ba^\top = b^\top \otimes a$.

It easy to see: \begin{align} (a^\top \otimes b) = \sum_{i=1}^n \sum_{j=1}^m a_ib_j e_{m,j} e_{n,i}^\top = \left(\sum_{j=1}^m b_j e_{m,j}\right)\left(\sum_{i=1}^n a_i e_{n,i}^\top\right) = ba^\top \end{align}

From there, $ba^\top \otimes ab^\top = a^\top \otimes b \otimes b^\top \otimes a = a^\top \otimes bb^\top \otimes a$.