I've been told that $u_{j} \frac{\partial u_{i}}{\partial x_{j}}=\underline{u} \cdot \underline{\nabla} \ \underline{u}$ and $u_{j} \frac{\partial u_{j}}{\partial x_{i}}=\frac{1}{2}\underline{\nabla} [\underline{u} \cdot \underline{u}]$
Can anyone explain why this is the case? I am aware that $\frac{\partial u_{i}}{\partial x_{j}}$ is the Hessian matrix, but I'm not sure how this leads to the result. Any advice?
Write $\underline{u} = (u_1, \ldots, u_n)$ where each $u_i$ is a function of $(x_1, \ldots, x_n)$. In the expression $u_j \frac{\partial u_i}{\partial x_j}$ there is an implicit summation on $j$ (this is called Einstein's summation convention ) so in fact
$$ u_j \frac{\partial u_i}{\partial x_j} = \sum_{j=1}^n u_j \frac{\partial u_i}{\partial x_j} .$$
This expression depends only on $i$ and you interpret it as the i-th coordinate of a vector:
$$ \left( u_j \frac{\partial u_1}{\partial x_j}, \ldots, u_j \frac{\partial u_n}{\partial x_j} \right) = \left( \sum_{j=1}^n u_j \frac{\partial u_1}{\partial x_j}, \ldots, \sum_{j=1}^n u_j \frac{\partial u_n}{\partial x_j} \right). $$
Now, $\underline{\nabla} \underline{u}$ is the differential of $\underline{u}$, a $n \times n$ matrix whose $(i,j)$ entry is $\frac{\partial u_j}{\partial x_i}$. If you interpret $\underline{u} \cdot (\underline{\nabla} \underline{u})$ as the product of a $1 \times n$ vector $\underline{u}$ by a $n \times n$ matrix you can see that $\underline{u} \cdot (\underline{\nabla} \underline{u})$ is precisely the vector $u_j \frac{\partial u_i}{\partial x_j}$.