derivative of tensor Hi I am trying to simplify
$$
A=\frac{1}{2}\left(\partial_j u_i+\partial_i u_j\right)\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}-\frac{2}{3}\delta_{ij}\delta_{kl}\right)\frac{1}{2}\left(\partial_l u_k+\partial_k u_l\right)
$$
where $\delta_{ij}$ is the usual Kronecker delta tensor, $\delta_{ij}=1$ for $i=j$ and $0$ otherwise. Note, I am specifically trying to simplify it using $\partial_i u_i=0$.  How can we do this?  My answer doesn't match the correct one.
  My attempt was to first calculate
$$
\frac{1}{2}\left(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}-\frac{2}{3}\delta_{ij}\delta_{kl}\right)\left(\partial_l u_k+\partial_k u_l\right)=\partial_j u_i+\partial_i u_j.
$$
Now we can write $A_{ij}$ as 
$$
A=\frac{1}{2}\left(\partial_j u_i+\partial_i u_j\right)\cdot (\partial_j u_i+\partial_u u_j)=\frac{1}{2} \left(\partial_j u_i +\partial_i u_j\right)^2
$$
Expanding the square I obtained
$$
A=\frac{1}{2} \left(\partial_j u_i \partial_j u_i +\partial_i u_j \partial_i u_j+ 2\partial_i u_j \partial_j  u_i\right)=\frac{1}{2}\left(\partial_j u_i \partial_j u_i +\partial_i u_j \partial_i u_j\right)+\partial_i u_j \partial_j  u_i
$$
The answer to this problem is supposed to be
$$
A= \left(\partial_j u_i \partial_j u_i +\partial_j u_i \partial_i u_j\right)
$$
which doesn't seem to be symmetric in $i,j$,
You can see this answer has three terms of the form $\partial_j u_i$, and only 1 term of the form $\partial_i u_j$.   How can I simplify my result to get this answer?  THanks.
Note
$$
\partial_j u_i\equiv \frac{\partial u_i}{\partial x_j}
$$

You seem to have figured it all out so here is just a quick answer to your final question.
The particular names we use for the summation variable(s) are just labels so we are free to change $(i,j)\to(j,i)$. Performing such a change gives us that
$$\partial_i u_j\partial_i u_j = \partial_j u_i\partial_j u_i$$ 
which is what you need to get to the final answer. This trick off relabeling the summation variables is a key ingedient to solving many tensor calculus problems so it's a good idea to try to remember it.
