I'm trying to figure out the whether the quadratic dual of the following algebra is anti-commutative.
My algebra is $A=T(V)/J$ where $J=\langle I\rangle$ and $I\subseteq\bigwedge^2(V)$. Firstly I think it follows simply by counting dimensions ($V$ is finite dimensional here) that $S^2(V)\subseteq J^{\perp}$. However, is this enough to say that the quadratic dual is anti-commutative, is there more to prove, or is it false?