We have two symmetric positive semidefinite matrices $A, B \in \mathbb{R}^{n\times n}$, with $\mbox{rank} (A) = r$ and $A = VV^T$ for some $V \in \mathbb{R}^{n\times r}$. Also, we know that $\mbox{ker} A \subseteq \mbox{ker} B$. Prove that $B = VQV^T$ for some $Q \in \mathbb{R}^{r\times r}$.
I can prove that $kerA = kerV^T$. So, $Ax = 0 = VV^Tx$. Hence $x^TVV^Tx = (V^Tx)^T(V^Tx) = 0$ and $V^Tx = 0$. Maybe it can helps somehow. Now it is obviuosly that if $B = VQV^T$ then $ker A \subseteq ker B$. But unfortunately I have to prove this is another direction.
Hint: There is a $U$ such that $B=UU^T$. If you can find an $M$ such that $VM=U$, then $Q=MM^T$.