Divergence of matrix-vector product 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?
 A: Let us write the matrix-vector product ${\bf M}\cdot {\bf c}$ in index notation (Einstein convention). 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
$$
\nabla\cdot ({\bf c}\cdot{\bf M}) = c_{i,j} M_{ij} + c_i M_{ij,j} = {\bf c}\cdot(\nabla\cdot {\bf M}) + {\bf M} : \nabla{\bf c} \, .
$$
A: 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, 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.
A: 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.
