For a matrix $A \in \mathbb R^{m\times n}$ with columns $a_i \in \mathbb R^m$, we know that $A^\top A$ is Positive Semidefinite (PSD).
I am interested in checking if B is positive semidefinite where $B_{ij} = (a_i^\top a_j)^2$. Each component of $B$ is the square of corresponding component of $A^\top A$.
Yes. This is a direct consequence of the Schur product theorem: the elementwise product of two positive semidefinite matrices is also positive semidefinite.