If Same Rank, Same Null Spaces? "If matrices B and AB have the same rank, prove that they must have the same null spaces."
I have absolutely NO idea how to prove this one, been stuck for hours now. Even if you don't know the answer, any help is greatly appreciated.

Let $\vec x\in ker(B)$ then $B\vec x=\vec 0$, and so $AB\vec x=A(B\vec x)=A(\vec 0) = \vec 0$.  Thus $\vec x \in ker(AB)$, so $ker(B)\subseteq ker(AB)$.   Now, $rank(B)=rank(AB)$, so by the rank-nullity theorem, $null(B)=null(AB)$.  Now $ker(B)\subseteq ker(AB)$, but $dim(ker(B))=dim(ker(AB))$, thus $ker(B)=ker(AB)$.
