$$
\newcommand{\Vec}{\operatorname{vec}}
\newcommand{\T}{^\top}
\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}
\newcommand{\R}{\mathbb{R}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\bmat}[1]{\begin{bmatrix} #1 \end{bmatrix}}
\newcommand{\pd}[2][]{\frac{\partial #1}{\partial #2}}
\newcommand{\normal}{\mathcal{N}}
\newcommand{\kron}{\otimes}
$$

# Kronecker Product and Matrix Vectorization
I demand that matrix multiplication take precedence over kronecker product.

## Definitions
- $\F$ is some field. In fact many results can be generalized to commutative rings.
- $e_i \in \F^n$ is the unit vector with the $i$ coordinate being one.
- $A \kron B \in \F^{(mp) \times (nq)}$ such that $(A\kron B)_{pi+k,qj+l} = A_{ij}B_{kl}$ where $A \in \F^{m \times n}$ and $B \in \F^{p \times q}$. Without ambiguity, its $ijkl$ entry is also denoted by $(A\kron B)_{ik,jl}$.
- $\Vec(A) = \sum_{j=1}^n e_j \kron Ae_j$ for $A \in \F^{m\times n}$.

## Properties

### Theorem
$(A \kron B)(C \kron D) = AC \kron BD$ for compatible matrices.
### Proof
$$
((A \kron B)(C \kron D))_{ik,jl}
= \sum_{p,q} (A \kron B)_{ik,pq} (C \kron D)_{pq,jl}
= \sum_{p,q} A_{ip} B_{kq} C_{pj} D_{ql}
\\ = \sum_p A_{ip} C_{pj} \sum_q B_{kq} D_{ql}
= (AC)_{ij} (BD)_{kl}
= (AC \kron BD)_{ik,jl}
$$

### Theorem
$\Vec(ABC) = (C\T \kron A)\Vec(B)$.
### Proof
Let $C \in \F^{m\times n}$.
$$
\begin{align*}
\Vec(ABC)
&= \sum_{j=1}^n e_j \kron ABCe_j
= \sum_{j=1}^n e_j \kron AB \sum_{i=1}^m C_{ij}e_i
= \sum_{i=1}^m \sum_{j=1}^n C_{ij}e_j \kron ABe_i
\\ &= \sum_{i=1}^m C\T e_i \kron ABe_i
= (C\T \kron A) \sum_{i=1}^m e_i \kron Be_i
= (C\T \kron A) \Vec(B)
\end{align*}
$$