Dot product between a vector and a matrix Suppose $v$ is a vector valued function. Then $\nabla v$ is a rank 2 tensor, i.e. a matrix. In the Navier-Stokes equations we see a term of the form $$(v \cdot \nabla) v.$$ At first I thought that the dot would indicate matrix multiplication, but as it's written, this does not work due to dimensions. How is one supposed to interpret/evaluate this term?
Also, what does the divergence of this term look like, i.e. $\nabla\cdot\big((v \cdot \nabla) v\big)$? Does it follow the usual product rule?
 A: Think about partial derivatives. The derivatives $\partial/\partial x_1$ and $\partial/\partial x_2$, for example, are derivative in the $x_1$ direction and in the $x_2$ direction, and I assume you have no problem with an expression like $\partial v/\partial x_1$. But now we see that $\partial/\partial x_2 = e_1\cdot\nabla$ and $\partial/\partial x_2 = e_2\cdot\nabla$, where we are simply forming differential operators and not differentiating anything in these expressions. (Of course, here $e_1, e_2$ are part of the standard basis.) This line of reasoning gives you $v\cdot\nabla$ as the derivative in the $v/|v|$ direction scaled by $|v|$. I would really think of this as the coordinate-free version of partial derivatives. In fact, for any totally-differentiable function $F(x)$, it's total derivative at $x$ can be expressed as the linear map
$$
  DF_x(v) = (v\cdot\nabla)F(x),
$$
and the matrix of this map in the standard basis is the Jacobian matrix.
The same way $\partial v/\partial x = (e_1\cdot\nabla)v$ is valid, so is $(v\cdot\nabla)v$; this is "the derivative of $v$ in the $v$ direction (scaled by $|v|$)". It's important to note here that I am fairly certain the intent is not to differentiate the $v$ in $v\cdot\nabla$; doing so is valid, but is harder to interpret.
And that brings us to your last question. These sorts of expressions are much easier to reason about when you're explicit about what's being differentiated. For example, I would write $(v\cdot\dot\nabla)\dot v$ to indicate that the right-hand $v$ is the only one being differentiated. This makes it easy to evaluate your divergence expression:
$$\begin{aligned}
  \nabla\cdot((v\cdot\nabla)v)
    &= \dot\nabla\cdot((\dot v\cdot\hat\nabla)\hat v) + \dot\nabla\cdot((v\cdot\hat\nabla)\dot{\hat v})
\\
    &= (\dot v\cdot\hat\nabla)(\dot\nabla\cdot\hat v) + (v\cdot\hat\nabla)(\dot\nabla\cdot\dot{\hat v})
\\
    &= (\dot v\cdot\hat\nabla)(\dot\nabla\cdot\hat v) + (v\cdot\nabla)(\nabla\cdot v),
\end{aligned}$$
where the hats and dots indicate which derivative is acting on what.
In the last line, I've made the differentiation implicit in the right-hand term since it follows usual conventions. This term is also easy to interpret: it's the $v$-directional derivative of the divergence of $v$. The left-hand term is stranger, a sort of "cross-divergence"; in coordinates
$$
  (\dot v\cdot\hat\nabla)(\dot\nabla\cdot\hat v)
    = \sum_{i,j}\frac{\partial v_i}{\partial x_j}\frac{\partial v_j}{\partial x_i}.
$$
