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.
**Hint:**

As the divergence is simply the sum of $n$ partial derivatives, I will show you how to deal with these derivatives.

If you have a matrix valued function $A$ and a vector valued function $\def\b{\mathbf}\b v$, then their product can be differentiated in the following way:

\begin{align}
\def\d{\partial}
\def\dt{\d t}
\def\div{\frac\d\dt}
\def\divp#1{\frac{\d #1}\dt}
\div A(t)\b v(t)
&=\div\sum\_{i,j}\b e\_iA\_{ij}(t)v\_j(t)\\
&=\sum\_{i,j}\b e\_i\left[v\_j(t)\divp{A\_{ij}(t)}+A\_{ij}(t)\divp{v\_j(t)}\right]\\
&=\sum\_{i,j}\b e\_i v\_j \divp{A\_{ij}(t)}+\sum\_{i,j}\b e\_i A\_{ij}(t)\divp{v\_j(t)}\\
&=\divp{A(t)}\b v(t)+A(t)\divp{\b v(t)}
.
\end{align}

You see it works like the usual product rule. I hope this helps you to find the final formula for the divergence.