 Quadratic ellipticity does not bound a nonsymmetric coefficient matrix

The quadratic form of a matrix only sees its symmetric part. Thus $\lambda |\xi {|}^2\le {\xi }^{T}A\xi \le \Lambda |\xi {|}^2$ need not bound the skew part of a measurable coefficient matrix. For $A=\left(\begin{array}{cc}1& k\\ -k& 1\end{array}\right)$ the quadratic form is exactly $|\xi {|}^2$, regardless of $k$. On a square, take $k(x)=1/|x_1|$ off the line $x_1=0$, with a finite arbitrary value on that line. Two compactly supported smooth functions that equal $x_2$ and $x_1$ near the origin have $(A\nabla u)\cdot \nabla v=k$ there; the purported energy pairing is not integrable. A Lax-Milgram theorem proof therefore requires bounded full coefficients or another hypothesis that makes the bilinear form bounded. Symmetry supplies the missing matrix bound from the quadratic upper bound.