If $Q_{ij}={1\over N} \sum_{\alpha} (u_i^\alpha u_j^\alpha -{1 \over 3} \delta_{ij})$, we have a another tensor (specifically the derivative of the free energy in a liquid crystal) given as the following $$aQ-b(Q^2- \frac{1}{3} Id|Q^2|)+cQ|Q^2|$$ where the term $$- \frac{1}{3} Id|Q^2|$$ is given as a Lagrange multiplier to enforce trace freeness. How does this make the whole thing trace free? I don't know if I fully understand Lagrange multipliers.
In this case the identity matrix is a 3x3 matrix. Note that $Q$ has the properties of $Q=Q^T$, $Tr(Q)=0$.
Thanks