Question: Suppose that $A$ is a matrix field and that $v$ is a vector field. What is the divergence of the matrix-vector product $A \cdot v$, which is a vector field?

Answer: 
I agree with Tommaso Seneci. This question deserves a better answer. Yes, it is just vector calculus, but there are some non-trivial tricks that deserve to be noted.

Inspired by [this note by Piaras Kelly](http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/Chapter_1_Vectors_Tensors/Vectors_Tensors_14_Tensor_Calculus.pdf), I can write down that 
$$
\nabla \cdot (\mathbf{A}\mathbf{v}) = (\nabla \cdot \mathbf{A}) \mathbf{v} + \text{tr}(\mathbf{A}\text{grad}\mathbf{v})
$$
where 
$$
\text{grad}\mathbf{v} =
\begin{pmatrix}
\frac{\partial v\_1}{\partial x\_1} & \frac{\partial v\_1}{\partial x\_2} & \frac{\partial v\_1}{\partial x\_3} \\
\frac{\partial v\_2}{\partial x\_1} & \frac{\partial v\_2}{\partial x\_2} & \frac{\partial v\_2}{\partial x\_3} \\
\frac{\partial v\_3}{\partial x\_1} & \frac{\partial v\_3}{\partial x\_2} & \frac{\partial v\_3}{\partial x\_3} \\
\end{pmatrix}
$$
and
$$
\nabla \cdot \mathbf{A} =
[\frac{\partial}{\partial x\_1} \quad \frac{\partial}{\partial x\_2} \quad \frac{\partial}{\partial x\_3}] \mathbf{A}
=
\begin{pmatrix}
\frac{\partial A\_{11}}{\partial x\_1}+\frac{\partial A\_{21}}{\partial x\_2}+\frac{\partial A\_{31}}{\partial x\_3} \\
\frac{\partial A\_{12}}{\partial x\_1}+\frac{\partial A\_{22}}{\partial x\_2}+\frac{\partial A\_{32}}{\partial x\_3} \\
\frac{\partial A\_{13}}{\partial x\_1}+\frac{\partial A\_{23}}{\partial x\_2}+\frac{\partial A\_{33}}{\partial x\_3} \\
\end{pmatrix}^T .
$$

The trick to do this calculation is this formula
$$
\nabla \cdot \mathbf{v} = \text{tr}(\text{grad}\mathbf{v}).
$$

First compute $\text{grad}(\mathbf{A}\mathbf{v})$ by product rule:
$$
\text{grad}(\mathbf{A}\mathbf{v})
=
[(\frac{\partial}{\partial x\_1} \mathbf{A})\mathbf{v} \quad (\frac{\partial}{\partial x\_2} \mathbf{A})\mathbf{v} \quad (\frac{\partial}{\partial x\_3} \mathbf{A})\mathbf{v}]
+
\mathbf{A} \text{grad}(\mathbf{v})
$$
Then take trace of the two terms. The trace of first term, by carefully simplifying, becomes $(\nabla \cdot \mathbf{A})\mathbf{v}$.

Please correct me if there is any mistake in the calculation.
Let us write the matrix-vector product ${\bf M}\cdot {\bf c}$ in index notation ([Einstein convention](https://en.wikipedia.org/wiki/Einstein_notation)). Using the product rule, the divergence of $({\bf M}\cdot {\bf c})\_{i} = M\_{ij} c\_j$ satisfies
$$
\nabla\cdot({\bf M}\cdot {\bf c}) = M\_{ij,i} c\_j + M\_{ij} c\_{j,i} = {\bf c}\cdot\left(\nabla\cdot({\bf M}^\top)\right) + {\bf M}^\top\! : \nabla{\bf c}\, ,
$$
where ${\bf A}:{\bf B} = \text{tr}({\bf A}^\top\!\cdot{\bf B}) = \text{tr}({\bf A}\cdot{\bf B}^\top)$.
Similarly, one shows that the vector-matrix product $({\bf c}\cdot {\bf M})\_{j} = c\_i M\_{ij}$ satisfies
$$
\nabla\cdot ({\bf c}\cdot{\bf M}) = c\_{i,j} M\_{ij} + c\_i M\_{ij,j} = {\bf c}\cdot(\nabla\cdot {\bf M}) + {\bf M} : \nabla{\bf c} \, .
$$