how to define the divergence operator of a matrix? This question related to my previous question here
For vector-valued function $u(x_1,x_2)=(u_1(x_1,x_2),u_2(x_1,x_2))$ we represents  the symmetric gradient $\mathcal E$ as follows
$$
\mathcal E(u)=
\begin{bmatrix}
\partial_1 u_1 & \frac12(\partial_2 u_1+\partial_1 u_2)\\
\frac12(\partial_2 u_1+\partial_1 u_2) & \partial_2 u_2
\end{bmatrix}
$$
my question: how to define the divergence of $\mathcal E(u)$? should $div(\mathcal E(u))$ be a 2 by 2 matrix or a 2 by 1 vector?
 A: Conventionally, divergence of a matrix is defined as the divergence of each column of this matrix. For example,
$$
A=\left(\mathbf{a}_1,\mathbf{a}_2,\cdots,\mathbf{a}_n\right),
$$
where $\mathbf{a}_j$ denotes the $j$-th column of the matrix $A$. Then
$$
\nabla\cdot A:=\left(\nabla\cdot\mathbf{a}_1,\nabla\cdot\mathbf{a}_2,\cdots,\nabla\cdot\mathbf{a}_n\right).
$$
However, this convention is sometimes challenged by other conventions. Take the Navier-Stokes equation for instance (where your matrix is exactly the viscosity tensor therein):
$$
\rho\left(\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla\right)\mathbf{v}=-\nabla p+\mu\nabla\cdot\mathbb{T}+\rho\mathbf{g},
$$
where $\mathbf{T}$ is a symmetric matrix. Vectors $\mathbf{v}$ and $\mathbf{g}$ are, by default, column vectors. But if you follow the definition above, $\nabla\cdot\mathbb{T}$ appears to be a row vector. Therefore, the convention in Navier-Stokes equation is that, after you figure out $\nabla\cdot\mathbb{T}$ as per the definition above, you need to transpose your result to make it a column vector.
