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According to Wikipedia:

The divergence of a continuously differentiable tensor field $\underline{\underline{\epsilon}}$ is:

$$\overrightarrow{\operatorname{div}}\,(\mathbf{\underline{\underline{\epsilon}}}) = \begin{bmatrix} \frac{\partial \epsilon_{xx}}{\partial x} +\frac{\partial \epsilon_{xy}}{\partial y} +\frac{\partial \epsilon_{xz}}{\partial z} \\ \frac{\partial \epsilon_{yx}}{\partial x} +\frac{\partial \epsilon_{yy}}{\partial y} +\frac{\partial \epsilon_{yz}}{\partial z} \\ \frac{\partial \epsilon_{zx}}{\partial x} +\frac{\partial \epsilon_{zy}}{\partial y} +\frac{\partial \epsilon_{zz}}{\partial z} \end{bmatrix} $$

How do you get this formula from the definition of divergence? Either formally, or with some abuse of notation?

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What is your definition of divergence? – Henning Makholm Nov 4 '11 at 2:03
Also you can work it out from the definition of $\nabla$ notation. – user13838 Nov 4 '11 at 2:54
up vote 1 down vote accepted

If S a matrix, with columns $S^{j}$, $j=1$, $n$ then $\mathrm{div}(S)_{j} = \mathrm{div}(S^{j})$.

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+1. More generally, for any linear operator that acts on vectors, the natural extension to matrices is to apply the operator column-wise. – user7530 Nov 10 '11 at 22:02

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