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What is $\nabla \cdot \mathbf{A}$ when $\mathbf{A} \in \mathbb{R}^{m \times n}$ is a matrix, and where is there a consise definition of this notation?

The Euler equations on Wikipedia contain terms on the form $\nabla \cdot (\mathbf{u} \otimes \mathbf{u} - w\mathbf{I})$ where $\mathbf{I}$ is the identity matrix. Other material on the Euler equations simply writes $\nabla \cdot (\mathbf{u} \mathbf{u} - w\mathbf{I})$. I assume $\mathbf{u} \otimes \mathbf{v} = \mathbf{u}\mathbf{v}^T \in \mathbb{R}^{m \times n}$ for $\mathbf{u} \in \mathbb{R}^m$ and $\mathbf{v} \in \mathbb{R}^n$.

Is $\nabla \cdot \mathbf{A}$ defined when $\mathbf{A}$ is not square / not symmetrical?

Svaberg
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    Your interpretation of $\otimes$ as outer product is correct, and indeed the Wikipedia page mentions this within the 'derivation of the conservation form' section (which needs to be clicked to be revealed). The use of Einstein notation within that section may further clarify the definitions used. – Semiclassical May 05 '17 at 14:11
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    So $\nabla \cdot \mathbf{A}, \ \mathbf{A} \in \mathbb{R}^{m \times n}$ is a column vector in $\mathbb{R}^n$ with elements $\sum_i \partial_i A_{ij}$, or is it a column vector in $\mathbb{R}^m$ with elements $\sum_j \partial_j A_{ij}$? The section appears to allude to a general definition of $\nabla \cdot \mathbf{A}$ but the definition is not given. – Svaberg May 05 '17 at 14:23
  • It should, I think, be the same interpretation as in this previous question. But without a specific definition it's hard to be sure. – Semiclassical May 05 '17 at 14:35

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Clicking "show" next to "Demonstration of the conservation form" reveals more context: the matrix $$\begin{pmatrix}{\mathbf {u}}\otimes {\mathbf {u}}+w{\mathbf {I}}\\{\mathbf {u}}\end{pmatrix}$$ is written out in components as $$ {\begin{pmatrix}u_{1}^{2}+w&u_{1}u_{2}&u_{1}u_{3}\\u_{2}u_{1}&u_{2}^{2}+w&u_{2}u_{3}\\u_{3}u_{1}&u_{3}u_{2}&u_{3}^{2}+w\\u_{1}&u_{2}&u_{3}\end{pmatrix}}$$ Other formulas clarify that $\nabla$ is applied to this matrix by differentiating the first column with respect to $x_1$, the second with respect to $x_2$, the third with respect to $x_3$, and then adding the results. The result being $$ \nabla\cdot {\begin{pmatrix}u_{1}^{2}+w&u_{1}u_{2}&u_{1}u_{3}\\u_{2}u_{1}&u_{2}^{2}+w&u_{2}u_{3}\\u_{3}u_{1}&u_{3}u_{2}&u_{3}^{2}+w\\u_{1}&u_{2}&u_{3}\end{pmatrix}} =\begin{pmatrix} (\operatorname{div} \mathbf u)\mathbf u + \operatorname{grad} \mathbf w \\ \operatorname{div} \mathbf u \end{pmatrix} $$ where I used explicit names to avoid any further $\nabla$-confusion.

If we think of $\nabla$ as a symbolic vector of partial derivatives, $\nabla = \begin{pmatrix} \partial /\partial x_1 \\ \partial /\partial x_2 \\ \partial /\partial x_3 \end{pmatrix}$, then the above is more properly $$ \begin{pmatrix}{\mathbf {u}}\otimes {\mathbf {u}}+w{\mathbf {I}}\\{\mathbf {u}}\end{pmatrix} \nabla = {\begin{pmatrix}u_{1}^{2}+w&u_{1}u_{2}&u_{1}u_{3}\\u_{2}u_{1}&u_{2}^{2}+w&u_{2}u_{3}\\u_{3}u_{1}&u_{3}u_{2}&u_{3}^{2}+w\\u_{1}&u_{2}&u_{3}\end{pmatrix}} \begin{pmatrix} \partial /\partial x_1 \\ \partial /\partial x_2 \\ \partial /\partial x_3 \end{pmatrix} $$ This makes sense for any matrix where the number of columns matches the dimension.