[example: The Laplacian]
Let $L=-\Delta$ on an [open set](/page/Open%20Set) $U\subsetneq\mathbb R^n$. In nondivergence form this means the second-order coefficient matrix is the identity matrix, so
\begin{align*}
a_{ij}&=\delta_{ij},\\
Lu&=-\sum_{i,j=1}^n \delta_{ij}\partial_{x_i}\partial_{x_j}u
=-\sum_{i=1}^n \partial_{x_i}\partial_{x_i}u
=-\sum_{i=1}^n\partial_{x_i}^2u.
\end{align*}
For every $\xi=(\xi_1,\dots,\xi_n)\in\mathbb R^n$, the principal quadratic form is
\begin{align*}
\sum_{i,j=1}^n\delta_{ij}\xi_i\xi_j
&=\sum_{i=1}^n \delta_{ii}\xi_i\xi_i+\sum_{\substack{1\le i,j\le n\\ i\ne j}}\delta_{ij}\xi_i\xi_j\\
&=\sum_{i=1}^n \xi_i^2+\sum_{\substack{1\le i,j\le n\\ i\ne j}}0\cdot \xi_i\xi_j\\
&=\sum_{i=1}^n \xi_i^2\\
&=|\xi|^2.
\end{align*}
Thus
\begin{align*}
\sum_{i,j=1}^n\delta_{ij}\xi_i\xi_j&\ge 1\cdot |\xi|^2
\end{align*}
for every $\xi\in\mathbb R^n$, so $-\Delta$ is uniformly elliptic with ellipticity constant $1$.
In divergence form the principal part is
\begin{align*}
-\sum_{i,j=1}^n\partial_{x_i}\left(\delta_{ij}\partial_{x_j}u\right)
&=-\sum_{i=1}^n\partial_{x_i}\left(\partial_{x_i}u\right)
=-\sum_{i=1}^n\partial_{x_i}^2u.
\end{align*}
With no first-order or zero-order terms, the associated [bilinear form](/page/Bilinear%20Form) satisfies, for $u\in H_0^1(U)$,
\begin{align*}
B[u,u]
&=\int_U\sum_{i,j=1}^n\delta_{ij}\partial_{x_j}u\,\partial_{x_i}u\,d\mathcal L^n\\
&=\int_U\sum_{i=1}^n \partial_{x_i}u\,\partial_{x_i}u\,d\mathcal L^n\\
&=\int_U\sum_{i=1}^n |\partial_{x_i}u|^2\,d\mathcal L^n\\
&=\int_U|\nabla u|^2\,d\mathcal L^n.
\end{align*}
Thus the matrix test and the variational test say the same thing: the identity principal matrix charges every coordinate direction, and the energy records exactly the full squared gradient.
[/example]