Trace of symmetric matrix product
Solution 1:
Since $B$ is rank one and positive-semi-definite (has to be p.s.d. and not p.d. since it is rank deficient) matrix, you have $B = u u^{T}$ for some $u \neq 0$. And so using $\mathrm{tr}(Auu^T) =\mathrm{tr}(u^TAu) = 0$, it follows that $u$ is an isotropic vector of $A$.
EDIT: Thanks to Loup Blanc for pointing out the mistake.