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The operator is strictly elliptic when its symmetric principal matrix $A(x)=(a^{ij}(x))$ is positive definite at every point:It is a uniformly elliptic operator when some $\lambda >0$ satisfiesfor every $x$ and $\xi$. Here the least eigenvalue of the continuous matrix $A(x)$ is a positive continuous function on the compact set ${\overline{B}}_1$. It therefore has a positive minimum, which supplies $\lambda$ and proves uniform ellipticity.

For a real second-order principal coefficient matrix $A(x)=(a^{ij}(x))$, uniform ellipticity means that one constant $\lambda >0$, independent of $x$, satisfiesFor measurable coefficients the condition is imposed almost everywhere; for a classical pointwise operator it holds at every relevant point. The quadratic form depends only on the symmetric part of $A$. Many estimates also assume a finite uniform upper bound or bounded full coefficients; the lower bound alone does not supply them.

For example, $A=\mathrm{diag}(1,4)$ has lower bound one. On $0<x_1<1$, the matrix $A(x)=\mathrm{diag}(x_1^2,1)$ is positive definite at each point but has no positive uniform lower bound: test $\xi =(1,0)$ as $x_1\downarrow 0$. On the restricted strip $x_1>\delta >0$, a bound $\min ({\delta }^2,1)$ works. This distinguishes a strictly elliptic operator from a uniformly elliptic operator on the whole domain.

In a divergence-form elliptic operator, the same inequality controls the principal gradient energy. Together with bounded coefficients and the Poincaré inequality, it underlies coercivity and weak existence. A nonsymmetric coefficient matrix needs separate boundedness control, as shown by quadratic ellipticity does not bound a nonsymmetric coefficient matrix.^[ref]