= Solution
For the stated <uniaxial nematic order> normalization, the <nematic director> is an <eigenvector> of $Q$ with <eigenvalue> $\lambda$, and the two perpendicular <eigenvalues> are $-\lambda/2$. Therefore
$$
\operatorname{Tr}Q^2=\frac32\lambda^2,
\qquad \operatorname{Tr}Q^3=\frac34\lambda^3,
\qquad (\operatorname{Tr}Q^2)^2=\frac94\lambda^4.
$$
Using precisely the <Landau-de Gennes free energy> convention in the preceding solution gives
$$
\boxed{\bar a=\frac{3a}{4},\qquad \bar b=\frac{9b}{16},\qquad \bar c=\frac c4}.
$$
If the quartic invariant were instead normalized as $b\operatorname{Tr}Q^4/4$, its coefficient would be $\bar b=9b/32$; the physical predictions are unchanged after redefining $b$.
For a nonzero <global minimizer> of the <free energy>, compare opposite values of the scalar <nematic order parameter>:
$$
f(\lambda)-f(-\lambda)=2\bar c\lambda^3.
$$
The lower one has $\bar c\lambda<0$, so
$$
\boxed{\operatorname{sgn}\lambda_{\rm eq}=-\operatorname{sgn}\bar c\quad(\bar c\ne0)}.
$$
The claim concerns the stable ordered phase, not every metastable stationary point. For $\bar c=0$ the two signs are degenerate. For $\bar c\ne0$, equality of the ordered and isotropic <free energies>, together with stationarity, gives $\lambda_t=-\bar c/(2\bar b)$ and $\bar a_t=\bar c^2/(4\bar b)>0$. The finite jump $\lambda_t$ exhibits the <first-order phase transition> driven by the cubic invariant.
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