# Rank and null space proof

I need to consider the following: if matrices $B$ and $AB$ have the same rank, prove that they have the same null spaces.

So my logic here is first to show that $\text{Null}(B) \subset \text{Null}(AB)$. Then I will use the rank + nullity theorem accordingly.

Let $\vec x\in \text{null}(B)$. Then $$B\vec x=\vec 0,\ \ \ \text{ and }\ AB\vec x=A(B\vec x)=A(\vec 0) = \vec 0.$$

Hence $\vec x \in \text{null}(AB)$, so $\text{null}(B)\subseteq \text{null}(AB)$.
So, $\text{rank}(B)=\text{rank}(AB)$, and by the rank-nullity theorem, $\text{null}(B)=\text{null}(AB)$.

Now $\text{null}(B)\subseteq \text{null}(AB)$, but $\dim(\text{null}(B))=\dim(\text{null}(AB))$, thus $\text{null}(B)=\text{null}(AB)$.

Looking to get confirmation that my workings are correct.

• No problem. Your reasoning looks flawless to me. – Martin Argerami Oct 7 '16 at 2:24
• awesome thanks for the help – jh123 Oct 7 '16 at 2:25

If $U \subset V$ are finite dimensional subspaces and $\dim U = \dim V$ then $U=V$.

Since $\dim {\cal R} (AB) = \dim {\cal R} B$, we have $\dim \ker (AB) = \dim \ker B$ and clearly $\ker B \subset \ker AB$.