Proof of vector calculus relation I am having difficulty to see why this is true:
$$(P\nabla)\cdot (P\nabla)=\nabla \cdot (P\nabla)$$
where $P$ is a projection operator
 A: Assuming $P$ is an orthogonal projection, then $P$ is a matrix and $P^2=P\ ,\ P^T=P$.
One can use index notation to simplify the calculations of both expressions, using the fact that if $\vec e_i$ is the $i$-th vector of the canonical basis $(i\in\{1,2,3\})$, then
$$
\vec e_i\cdot\vec e_j=\delta_{ij}=
\left\{
\begin{array}{l}
1 \quad\text{if}\quad i=j\\
0 \quad\text{if}\quad i\neq j
\end{array}
\right.\quad .
$$
$$
\nabla=\sum_{i=1}^{3}\vec e_i\frac{\partial}{\partial x_i}
$$
$$
\left.
\begin{array}{c}
P_{ij}=P_{ji}\\
(P^2)_{ij}=\sum_{k}^3P_{ik}P_{kj}=P_{ij}
\end{array}
\quad
\right\}
\quad
P\ \text{is orthogonal projection}
$$
$$
P\nabla=\sum_{i,j=1}^{3}\vec e_iP_{ij}\frac{\partial}{\partial x_j}
$$
$$
\begin{array}{rcl}
(P\nabla)\cdot(P\nabla)&=&\sum_{i,j,k,l=1}^{3}\vec e_iP_{ij}\frac{\partial}{\partial x_j}\vec e_kP_{kl}\frac{\partial}{\partial x_l} \\
&=& \sum_{i,j,k,l=1}^{3}
\underbrace{\vec e_i\cdot\vec e_k}_{=\delta_{ik}}
P_{ij}P_{kl}\frac{\partial^2}{\partial x_j\partial x_l} \\
&=& \sum_{j,k,l=1}^{3}
\underbrace{P_{kj}P_{kl}}_{=P_{jk}P_{kl}}
\frac{\partial^2}{\partial x_j\partial x_l} \\
&=& \sum_{j,l=1}^{3}
\underbrace{P^2_{jl}}_{=P_{jl}}
\frac{\partial^2}{\partial x_j\partial x_l}
\end{array} 
$$
On the other hand,
$$
\begin{array}{rcl}
\nabla\cdot(P\nabla) &=& \sum_{i,j,k=1}^3
\underbrace{\vec e_i\cdot\vec e_j}_{=\delta_{ij}}
P_{jk}\frac{\partial^2}{\partial x_i\partial x_k} \\
&=& \sum_{i,k=1}^3P_{ik}\frac{\partial^2}{\partial x_i\partial x_k}=(P\nabla)\cdot(P\nabla)
\end{array}
$$
If you imagine $\nabla$ as being a vector in space, and $P\nabla$ as projecting $\nabla$ orthogonally in some direction, resulting a vector in that direction; then in the LHS of your equation you have the square of the length of that projection. The RHS is the scalar product of $\nabla$ and its own projection in that direction, which is also the square of the length of the projected $\nabla$.
A: This seems to be not true. You have $\nabla f=\sum \frac{\partial f}{\partial i}e_{i}$, so $P\nabla f=\frac{\partial f}{\partial i}$. Applying it twice gives you the second directional derivative at $e_{i}$. The right hand side is $\nabla \frac{\partial f}{\partial i}$, and there will be other terms involving mixed $ij$ terms. They do not need to be equal. 
