According to http://en.wikipedia.org/wiki/Kronecker_product
If we have linear operators $A,B$ representing linear maps $ S : V → X$ and $T : W → Y,$ then the kronecker product $A \otimes B $ has the property that for basis elements $v_i,w_j$ of $V$ and $W$ respectively then
$$A \otimes B(v_i \otimes w_j) = (Av_i)\otimes(Bw_j)$$
I am confused how the dimensions work out here. Let us limit this to the case where $V = X =W = Y = \mathbb{C}^n$. Then $A \otimes B$ would be a matrix of dimension $\mathbb{C}^{n^2,n^2}$ but wouldn't $v_i \otimes w_j$ be a matrix of dimension $\mathbb{C}^{n,n}$ and you can't multiply these matrices.
Perhaps my definition of tensor product is wrong, for $\mathbb{C}^n$ or $\mathbb{R}^n$ I believe the tensor product to be the outer product : http://en.wikipedia.org/wiki/Outer_product
Thanks for your help
Best Answer
That's all a question of indexing. Recall that $\mathbb C^{n,n}$ is just a vector space of dimension $n \cdot n = n^2.$ So $v_i\otimes w_j$ can very well be multiplied with a matrix from $\mathbb C^{n^2,n^2}.$
More technically, it works as follows. Let $B := \{e_j\ |\ 1 \leq j \leq n \}$ be a basis of $\mathbb C^n,$ the standard basis, if you want. Then $C:= \{e_j\otimes e_k\ |\ 1 \leq j,k \leq n \}$ is a basis of $\mathbb C\otimes\mathbb C.$ Then the general element $x$ of $\mathbb C\otimes\mathbb C$ can be written as $x = \sum_{j,k}\xi_{jk}e_j\otimes e_k,$ so its coefficient vector relative to the basis $C$ is $(\xi_{jk}).$ Note that we have two indices, not just one. You can view it as a matrix, as you remarked above.
Now, let $E_{jk} \in \mathbb C^{n,n}$ be the matrix with $1$ at the intersection of the $j$-th row and the $k$-th column and $0$ elsewhere. Then $D:= \{E_{jk}\ |\ 1\leq j,k\leq n\}$ is a basis of $\mathbb C^{n,n}.$ As above, $E:= \{E_{jk}\otimes E_{pq}\ |\ 1\leq j,k,p,q\leq n\}$ is a basis of $\mathbb C^{n,n}\otimes \mathbb C^{n,n}.$ The general element $M$ of $\mathbb C^{n,n}\otimes \mathbb C^{n,n}$ can be written as $M = \sum_{j,k,p,q}\mu_{jkpq}E_{jk}\otimes E_{pq}$ and has coefficient "matrix" $(\mu_{jkpq}).$ Here we have four indeces instead of the usual two.
With all this notation, the coefficient vector of $Mx$ is $$ \left(\sum_{k,q}\mu_{jkpq} \xi_{kq}\right) $$ with free indices $j$ and $p$. This is very similar to the "usual" matrix-vector-multiplication, except that each dimension has a pair of indices, not just one index. However, note the "intertwining" of indices. The indices of $\xi$ are connected to the second and fourth indices of $\mu,$ not the third and fourth.
Finally, let's consider the Kronecker products of $S = (\sigma_{jk}) \in \mathbb C^{n,n},$ $T = (\tau_{pq}) \in \mathbb C^{n,n},$ $x = (\xi_k) \in \mathbb C^n,$ and $y = (\eta_q) \in \mathbb C^n.$ We have $$ S\otimes T = \left(\sigma_{jk}\tau_{pq}\right) \in \mathbb C^{n,n}\otimes \mathbb C^{n,n} $$ and $$ x\otimes y = \left(\xi_k\eta_q\right) \in \mathbb C^n \otimes \mathbb C^n. $$ From this, we get $$ \begin{align} (S\otimes T)(x\otimes y) & = \left(\sum_{k,q}\sigma_{jk}\tau_{pq}\xi_k\eta_q\right) \\ & = \left(\left(\sum_k\sigma_{jk}\xi_k\right)\left(\sum_q\tau_{pq}\eta_q\right)\right) \\ & = (Sx)\otimes(Ty) \in \mathbb C^n \otimes \mathbb C^n. \end{align} $$ We find that everything fits nicely together.