Kronecker product and outer product confusion I have two column vectors:
\begin{equation}
u = 
\left[\matrix{
   1 \cr
   2\cr
}\right]
\end{equation} 
\begin{equation}
v = 
\left[\matrix{
   4 \cr
   4\cr
}\right]
\end{equation}
I'm trying to compute the Kronecker product of two vectors $u \otimes v$.
As I understand, the outer product of vectors is a special case of the Kronecker product of matrices.
http://en.wikipedia.org/wiki/Kronecker_product says:

If A is an m × n matrix and B is a p × q matrix, then the Kronecker
  product $A \otimes\ B$ is the mp × nq block matrix.


http://en.wikipedia.org/wiki/Outer_product says:



So will $u \otimes v$ be of dimension 4 × 1 (according to the first definition) or 2 × 2 (according to the second definition)?
 A: The relation between outer product $\circ$  and kronecker product $\otimes$ .
$${\textbf{a}}_{I \times 1} \otimes {\textbf{b}}_{J \times 1} = vec(({\textbf{a}}\circ {\textbf{b}})^T) = vec(({\textbf{a}}{\textbf{b}^T}_{I \times J})^T_{J \times I})_{JI \times 1}$$
Before taking vectorization you should do a transpose.
In case of 3 vectors, the resulted matrix as before is multiplied by each entry in c vector and each result is a frontal slice of a tensor. Notice that $.*$ is an element wise product between each element in $\textbf{c}$ vector and the resultant matrix of $(\textbf{a} \circ \textbf{b})$ and not the element wise product between the whole vector $\textbf{c}$ and $(\textbf{a} \circ \textbf{b})$.
$$ {\textbf{a}}_{I \times 1} \otimes {\textbf{b}}_{J \times 1} \otimes {\textbf{c}}_{K \times 1} = vec(( ({\textbf{a}}\circ {\textbf{b}}) .* {\textbf{c}})^T_{I \times J \times K})_{IJK \times  1}  $$ 
Outer product between 2, each of 2 dimension then the result would be 4 dimension tensor which is totally different than the kronecker product
Kronecker Product for matrices
$$ {\textbf{A}}_{I \times R} \otimes {\textbf{B}}_{ J \times R} = {\textbf{D}}_{ IJ \times R^2}  $$
$$ {\textbf{D}} = \begin{bmatrix}
    \vdots     & \vdots & & \vdots & & \vdots \\ 
    \textbf{a}_1 \otimes {\textbf{b}_1}  & {\textbf{a}_1} \otimes {\textbf{b}_2} & \dots &  \textbf{a}_1 \otimes {\textbf{b}_R} & \dots &\textbf{a}_R \otimes {\textbf{b}_R} \\ 
    \vdots     & \vdots & & \vdots & & \vdots
\end{bmatrix}$$
Outer Product for matrices
$$ \textbf{A}_{I \times R_1} \circ \textbf{B}_{J \times R_2} = {\textbf{C}}_{I \times J \times R_1 \times R_2}$$
to show how the result would be you have to do it column by column outer product,
$$ {\textbf{a}_1} \circ {\textbf{b}_1} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{x} & {\textbf{y}} \\ 
    \vdots     & \vdots
\end{bmatrix},  {\textbf{a}_1} \circ {\textbf{b}_2} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{m} & {\textbf{n}} \\ 
    \vdots     & \vdots
\end{bmatrix},{\textbf{a}_2} \circ {\textbf{b}_1} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{p} & {\textbf{q}} \\ 
    \vdots     & \vdots
\end{bmatrix},{\textbf{a}_2} \circ {\textbf{b}_2} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{j} & {\textbf{k}} \\ 
    \vdots     & \vdots
\end{bmatrix},$$
The final result would be,
$${\textbf{C}}_{I \times J \times 1 \times 1} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{x} & {\textbf{p}} \\ 
    \vdots     & \vdots
\end{bmatrix}, {\textbf{C}}_{I \times J \times 2 \times 1} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{y} & {\textbf{q}} \\ 
    \vdots     & \vdots
\end{bmatrix}, {\textbf{C}}_{I \times J \times 1 \times 2} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{m} & {\textbf{k}} \\ 
    \vdots     & \vdots
\end{bmatrix}, {\textbf{C}}_{I \times J \times 2 \times 2} = \begin{bmatrix}
    \vdots     & \vdots \\ 
    \textbf{n} & {\textbf{j}} \\ 
    \vdots     & \vdots
\end{bmatrix}$$
A: This is a very good example of abuse of notation, more precisely, reload of operator. Actually the operator $\otimes$ is usually used as tensor product, which is a bilinear operator. It's easy to verify that both Kronecker product (denoted by $\otimes_K$) and outer product (denoted by $\otimes_O$) are bilinear and special forms of tensor product. For example, given two vectors $u,v\in V$, we have
$$u\otimes_O v=u\otimes_Kv^H$$
This is why wiki says outer product is a special case of Kronecter product.
