Trouble expanding a del operator expression So when messing with some PDE, I came across this expression:
$$\nabla \cdot [(\vec{u} \cdot\nabla)\vec{u}] \hspace{12mm}[1]$$
I then tried to find whether I can expand it by breaking it down component by component and then reassemble it (assuming $\vec{u}=(u1,u2,u3)$, where the us are functiosn of x,y,z) and then I got this after 5 hours of algebra
$$\vec{u}\cdot\nabla^2\vec{u}+\nabla \vec{u} \cdot \nabla\vec{u} \hspace{12mm}[2]$$
I also tried to naviely use the vector identity given in wikipedia
$$\nabla \cdot (\psi\vec{A})=\vec{A} \cdot \nabla\psi+\psi\nabla \cdot \vec{A}$$
to get something like this
$$\nabla(\vec{u}\cdot\nabla)\cdot \vec{u} + (\vec{u} \cdot \nabla)\nabla \cdot \vec{u} \hspace{12mm}[3]$$
But after expanding [3] component by component and compare with [1], I found it is off by the presence of the 2nd term of [3]
I also noticed that if the following is true
$$\nabla \cdot [(\vec{u} \cdot\nabla)\vec{u}]=\vec{u}\cdot\nabla^2\vec{u}+\nabla \vec{u} \cdot \nabla\vec{u}$$
then by what we know about vector calculus, the tensorial order of this expression should be 0
However I have trouble interpreting this
$$\nabla \vec{u} \cdot \nabla\vec{u}$$
If I treat the vector gradient as a 2nd order tensor (e.g. a matrix), then this is like matrix multiplication, but this will give me a tensorial order of 2 and not 0
It I treat it instead as some sort of norm squared, then I am at lost on how to compute it (What type of matrix norm would this be?)
So basically

Q1. How to interpret this term?
$$\nabla \vec{u} \cdot \nabla\vec{u}$$
Q2. Why does
$$(\vec{u} \cdot \nabla)\nabla \cdot \vec{u}$$
have to vanish for [1] to be true. how to interpret this geometrically?

 A: Okay, first you've done a lot of voodoo here.  Let's try to simplify this from first principles.
First, consider the $u \cdot \nabla u$ part--take as convention that $u \cdot \nabla u = (u \cdot \nabla) u$ for the rest of this post.
We can expand this by an identity to
$$u \cdot \nabla u = \frac{1}{2} \nabla u^2 - u \times (\nabla \times u)$$
Now we can take a divergence:
$$\nabla \cdot (u \cdot \nabla u) = \frac{1}{2} \nabla^2 u^2 - \nabla \cdot [u \times (\nabla \times u)]$$
That second term is going to be fun.  Let's expand it:
$$\nabla \cdot [u \times (\nabla \times u)] = (\nabla \times u)^2 + u \cdot [\nabla \times (\nabla \times u)]$$
That should get you an identity involving just derivatives of $u$ or $u^2$.
Now, regarding your two questions:  for (Q1), I don't even see how that question can be answered.  If you don't have any definition of what a gradient of a vector field is, then how can you justify the manipulation that got you to write it down?
For (Q2), when you directly expand $\nabla \cdot (u \cdot \nabla u)$ according to the product rule, you get at the least
$$\dot \nabla \cdot (\dot u \cdot \nabla u) + (u \cdot \nabla)(\nabla \cdot u)$$
where the dots denote that $\dot \nabla$ differentiates only $\dot u$.  So the term you're concerned about is definitely there; it's possible it cancels in some way, but I suspect that the $\nabla \cdot (\psi A)$ identity you tried to use doesn't hold when $\psi$ is a scalar differential operator instead of a scalar field.
