= Quadratic ellipticity does not bound a nonsymmetric coefficient matrix
{title2=$\xi^TA\xi=\xi^T\tfrac12(A+A^T)\xi$}
The <quadratic form> of a matrix only sees its symmetric part. Thus $\lambda|\xi|^2\leq\xi^TA\xi\leq\Lambda|\xi|^2$ need not bound the skew part of a measurable coefficient matrix. For $A=\begin{pmatrix}1&k\\-k&1\end{pmatrix}$ 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.
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