= Solution
In the variational convention needed here, a <positive definite symmetric operator> on a real <Hilbert space> satisfies
$$
\langle Lu,v\rangle=\langle u,Lv\rangle,\qquad
\langle Lv,v\rangle>0\quad(v\ne0).
$$
For a bounded operator on the whole space, symmetry is <self-adjointness>. On a complex <Hilbert space>, the corresponding convention is Hermitian positivity; the quadratic expression is real. Strict positivity should be distinguished from <uniformly positive definite symmetric operator> coercivity, which requires a fixed $\gamma>0$ such that $\langle Lv,v\rangle\ge\gamma\|v\|^2$.
Symmetry matters in a real space. If “positive definite” means only the inequality for the <quadratic form>, the next variational assertion is false: $L=\begin{pmatrix}1&-1\\1&1\end{pmatrix}$ has $\langle Lv,v\rangle=\|v\|^2>0$, but its skew part disappears from that form. Its <Euler-Lagrange equation> is $v=f$, not $Lv=f$. Thus the usual symmetric/Hermitian convention, or an explicit symmetry assumption, is required.
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