# How to write a matrix $\mathcal{M}$ such that $\mathcal{M} \boldsymbol{x}=\boldsymbol{\omega}\times\boldsymbol{x}$? [duplicate]

As is well known, it is possible to use the $$\nabla$$ operator as if it were a vector.  Someone consider it an abuse of notation but surely something that works well and is very useful. Well, how is it possible to consider the operator $$\boldsymbol{\omega}\times$$ as a matrix? How build a matrix $$\mathcal{M}$$ such that $$\boldsymbol{\omega} \times \boldsymbol{x} = \mathcal{M} \boldsymbol{x}$$?

The present answer is already correct, just let me show how to get the result. Using Einstein convention (i.e. repeated indices are summed) the product $$c = a \times b$$ can be written as
$$c_i = (a \times b)_i =\epsilon_{ijk} \, a_j \, b_k = A_{ik} b_k$$
where $$\epsilon_{ijk}$$ is the Levi-Civita symbol and the matrix $$A$$ is defined by $$A_{ik}=\epsilon_{ijk} \, a_j=-\epsilon_{iks} \, a_s$$. It is easy to see that $$A$$ is skew-symmetric, i.e. $$A_{lm} =-A_{ml}$$. This trick will be useful to study the $$SO(3)$$ group and its associated algebra $$so(3)$$.
$$[\mathbf{a}]_{\times} = \begin{bmatrix} \,\,0 & \!-a_3 & \,\,\,a_2 \\ \,\,\,a_3 & 0 & \!-a_1 \\ \!-a_2 & \,\,a_1 & \,\,0 \end{bmatrix},$$