I accidentally found that curl can be represented as a matrix, and conformed it on wiki:
$$[\nabla ]_{\times}=\begin{bmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{bmatrix}$$
then something quite strange I am not sure
since for any cross product vector:
$A\times{(B}\times{C})$
is not always equal to
$(A\times{B})\times{C}$
but matrix can do this:
$
(MN)Q=M(NQ)
$
So at the end of the day, is curl or cross product actually matrix?
Best Answer
You are not dealing with two cross products. If ${\bf F} = A{\bf e}_1+B{\bf e}_2 + C {\bf e}_3$, then: $$\begin{bmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{bmatrix} \begin{bmatrix} A \\ B \\ C \end{bmatrix} = \begin{bmatrix} -B_z + C_y \\ A_z - C_x \\ -A_y + B_x \end{bmatrix}$$ is exactly the coordinates of $\nabla \times {\bf F}$ in the basis $\{ {\bf e}_i\}_{i=1}^3$. We check this: $$\begin{vmatrix} {\bf e}_1 & {\bf e}_2 & {\bf e}_3 \\ \partial_x & \partial_y & \partial_z \\ A & B & C \end{vmatrix} = (C_y - B_z){\bf e}_1 + (A_z - C_x){\bf e}_2 + (B_x - A_y){\bf e}_3.$$
To elaborate a bit more: the curl $\nabla\times$ is an operator that takes a vector field $\bf F$ and spits another one: $\nabla \times {\bf F}$. The matrix there is just a representation. If you write the curl, say, in spherical coordinates, you'll have another matrix with $\partial_r,\partial_\theta$ and $\partial_\phi $, that will do the same job: take the spherical coordinates of a vector field to the spherical coordinates of its curl.