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Considering that two vectors $A \times B$ = $\hat A* B$, where $\hat A$ is a skew symmetric matrix containing elements of $A$

Can we then write the curl $\nabla \times A$ as $\partial \vec r *A$ where $\partial \vec r$ = $[\partial x, \partial y, \partial z]$

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$\nabla \times \vec A$ can be written in terms of a tensor $T$ operator with

$$T = \hat x \hat y(-\partial_z)+\hat x \hat z(\partial_y)+\hat y \hat x(\partial_z)+\hat y \hat z(-\partial_x)+\hat z \hat x(-\partial_y)+\hat z \hat y(\partial_x)$$

Then, for all differentiable vector fields $\vec A$, we have

$$\nabla \times \vec A=T\cdot \vec A$$

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Another way to see this (implicit summation over repeated indices which label the three Cartesian coordinates $x,y,z$): $$(\nabla \times \mathbf{v})_i = \epsilon_{ijk} \partial_j v_k$$ Let $$T_{ik} = \epsilon_{ijk} \partial_j$$ which as a matrix looks like $$T_{ik} = \begin{pmatrix} 0 & -\partial_z & \partial_y \\ \partial_z & 0 & -\partial_x \\ -\partial_y & \partial_x & 0 \end{pmatrix} $$ then by construction, $$(\nabla\times \mathbf{v})_i = T_{ij} v_j$$

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