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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 $\boldsymbol{\omega} \times \boldsymbol{x} = \mathcal{M} \boldsymbol{x}$?

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2 Answers 2

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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 $so(3)$.

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$$ [\mathbf{a}]_{\times} = \begin{bmatrix} \,\,0 & \!-a_3 & \,\,\,a_2 \\ \,\,\,a_3 & 0 & \!-a_1 \\ \!-a_2 & \,\,a_1 & \,\,0 \end{bmatrix}, $$

Source: https://en.wikipedia.org/wiki/Skew-symmetric_matrix#Cross_product

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