Let $u \in H^1(\Omega) $ and $v \in H^1_0(\Omega) $ be 2 vectors
And let $D(u)= \frac{\nabla u + \nabla u^t}{2}$ , and since the gradient of a vector is a matrix then $D(u)$ is a matrix.
How can I prove that:
$\int_{\Omega} div(D(u)).v dx = - \int_{\Omega} D(u):D(v) dx $
We use integration by parts and we know that $v=0$ on the boundary of $\Omega$
I suppose your relation is wrong, I expect $$ \int_\Omega\operatorname{div}(D(u))v_idx $$ then, taking into account that
\begin{align} \frac{1}{2}(u_{i,j}+u_{j,i})_{,j}v_i &=\left[\frac{1}{2}(u_{i,j}+u_{j,i})v_i\right]_{,j}-\frac{1}{2}(u_{i,j}+u_{j,i})v_{i,j}=\\ &=\left[\frac{1}{2}(u_{i,j}+u_{j,i})v_i\right]_{,j}-\frac{1}{2}(u_{i,j}+u_{j,i})\frac{1}{2}(v_{i,j}+v_{j,i})= \end{align} where the last passage is due to the symmetry of the first factor, you arrive at the desired result.
This is just the integration by parts: \begin{align} \int_\Omega\operatorname{div}(D(u))v_idx&=\int_\Omega\left[\frac{1}{2}(u_{i,j}+u_{j,i})v_i\right]_{,j}dx-\int_\Omega\frac{1}{2}(u_{i,j}+u_{j,i})\frac{1}{2}(v_{i,j}+v_{j,i})dx\\ &=\int_{\partial\Omega}\left[\frac{1}{2}(u_{i,j}+u_{j,i})v_i\right]n_jda-\int_\Omega D(u):D(v)dx \end{align}