= Solution
A <nematic director> is an unoriented axis: $\mathbf n$ and $-\mathbf n$ describe the same molecular alignment. A vector average cannot represent this head-tail symmetry. The orientational second moment $\langle n_in_j\rangle$ is a <symmetric second-rank tensor>, and removing its isotropic part gives the <traceless second-rank tensor> $Q_{ij}\propto\langle n_in_j-\delta_{ij}/2\rangle$. It vanishes for an isotropic angular distribution and transforms correctly under <rotation matrices>.
Every nonzero real symmetric traceless $2\times2$ matrix has <eigenvalues> $\lambda/2,-\lambda/2$ and can be written $Q=\lambda(\mathbf n\mathbf n^{\mathsf T}-I/2)$ by choosing the eigenvector of the positive eigenvalue as director and $\lambda\ge0$. Consequently $\operatorname{Tr}Q^2=\lambda^2/2$. The bulk <Landau-de Gennes free energy> becomes
$$
f_{\rm bulk}=\frac a2\lambda^2+\frac b4\lambda^4=\frac b4\left(\lambda^2+\frac ab\right)^2-\frac{a^2}{4b}.
$$
For $a<0$, its minimum occurs at $\lambda^2=-a/b$. The elastic density $K|\nabla\cdot Q|^2/2$ is nonnegative and vanishes for a uniform field. Thus, with no <boundary condition> imposing a texture, the pointwise lower bound is attained by
$$
\boxed{\lambda_0=\sqrt{-a/b},\qquad Q_{ij}=\lambda_0(n_in_j-\delta_{ij}/2),\qquad f_{\min}=-a^2/(4b).}
$$
The negative-amplitude representation is equivalent in two dimensions to rotating the director by $\pi/2$, so the positive choice loses no tensor minima. This is minimization of the specified coarse-grained functional, not a claim of true thermodynamic long-range nematic order at every <temperature> in an infinite two-dimensional system. The PDF explicitly uses the divergence elastic term; the TeX's $|\nabla Q|^2$ loses the dot and would give different elastic coefficients.
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