Null space for $AA^{T}$ is the same as Null space for $A^{T}$

Let $q_i$ be a null vector of $A A^\top$, i.e. $ A A^\top q_i =0 $, then $ 0 = q_i^\top A A^\top q_i = \vert\vert A^\top q_i \vert\vert_2$, and thus $q_i$ is also a null vector of $A^\top$. Thus $\mathrm{rank}(A A^\top) = \mathrm{rank}(A^\top) = \mathrm{rank}(A)$.