Showing that the bilinear form is H-elliptic How do I show that the bilinear form 
$$a(u,u) = \int_\Omega A \nabla u \nabla u \, dx  $$
is H-elliptic?
I am not sure if I can directly say that $$a(u,u) \geq \alpha \lVert \nabla u \rVert^2_{L^2(\Omega)} = \alpha \lVert \dot{u}\rVert^2_{W^(\Omega)} \geq  \alpha \lVert u \rVert^2_{W^(\Omega)} .$$
I am working with the space $W(\Omega) = H^1(\Omega) / \mathbb{R}$, whose norm is 
$$\lVert \nabla u \rVert^2_{L^2(\Omega)} = \lVert \dot{u}\rVert^2_{W^(\Omega)},$$
where $\dot{u}$ is an equivalence class of $u$. Thank you.
 A: Edit: After looking at a more recent question of yours, I realise that I  missed that your $ W(\Omega) $ is a quotient space. In comments within this other post, you also explained that you are studying elliptic PDEs with Neumann boundary conditions, which is very important information!
I wrote my original answer under the assumption that you were dealing with Dirichlet boundary conditions, and I interpreted your final $|| u ||_{W(\Omega)}$ to symbolise $$ || u ||_{H^1(\Omega)} = \left( || u ||_{L^2(\Omega)}^2 + || \nabla u ||_{L^2(\Omega)}^2 \right)^{\frac 1 2},$$ which is the quantity relevant for the purposes of proving existence of weak solutions to to elliptic PDEs with Dirichlet boundary conditions. My original answer only applies for $$u \in H_0^1(\Omega) = \overline{C_{\rm comp. \ supp.}^\infty(\Omega)} \subset H^1 (\Omega),$$ which clearly isn't the correct space of functions if your motivation is to study Neumann boundary conditions.
I'll leave the original answer here for the time being, in case you want to compare notes. (In the comments within your other post, we discussed replacing the use of the Poincare inequality in this answer with Poincare-Wirtinger.)

Original answer: I presume the integral looks like this:$$a(u,u) = \int_\Omega  A_{ij}(x)   \nabla_i u \nabla_j u \ dx$$ where $\Omega \subset \mathbb R^n$ is a bounded open domain and $n \geq 3$.
I also presume $\alpha > 0$ is a constant such that $$\sum_{i,j}A_{ij}(x) v_i v_j \geq \alpha \sum_i v_i v_i \ \ \ \ \ \ \ \  {\rm for \ all \ \ } x \in \Omega, \ \  v_i \in \mathbb R^n.$$
So what you have showed is that
$$ a(u,u) \geq \alpha || \nabla u ||_{L^2 (\Omega)}$$
for all $u \in H_0^1 (\Omega)$.
To finish off, we need to use Poincare's inequality (which is a simple consequence of the Sobolev inequality) to deduce that there exists a $C > 0$ (independent of the choice of $u$, but dependent on the choice of $\Omega$) such that
$$ || u ||_{L^2(\Omega)} \leq C || \nabla u ||_{L^2(\Omega)} \ \ \ \ \ \ \ \ \ {\rm for \ all \ } u \in H^1_0(\Omega).$$
Therefore,
\begin{multline} a(u,u) \geq \alpha || \nabla u ||_{L^2 (\Omega)} \geq \frac{\alpha}{(1 +  C^2)^{\frac 1 2}} \left(|| u ||_{L^2(\Omega)}^2 + || \nabla u ||_{L^2(\Omega)}^2 \right)^{\frac 1 2} \\  = \frac{\alpha}{(1 +  C^2)^{\frac 1 2}} || u ||_{H^1 (\Omega)} \ \ \ \ \ 
{\rm for \ all \ \ } u \in H_0^1(\Omega).\end{multline}
