Since $\theta^2=Id$, its eigenvalues are 1 or -1. Suppose $\theta^2(u)=-u$, this implies that $-B(u,\theta(u))=B(u,u)<0$ since $B$ is negative definite. This is a contradiction since $B_{\theta}$ is definite positive. An involution which as only $1$ as eigenvalue is the identity.