# Show that $nullity(B)\leq nullity(AB)$

Let A be an $m\times k$ matrix; Let B be an $k\times n$ matrix.

a) Show that: $nullity(B)\leq nullity(AB)$

We had this in our previous midterms and even after viewing the official solution I couldn't keep up with it.

The official solution was as follows: $$\vec{x}\subset null(B)$$ $$B(\vec{x})=0$$ $$A(B(\vec{x}))=A(\vec{0})=0$$

Then, $dim(null(B))\leq dim (null(AB))$.

I don't understand at all how just by showing that $\vec{x}$ in the $nullity(AB)$ is enough to show that the dimensions is greater than $nullity(B)$.

Can someone please provide a reason as to why or a better way to prove this.

• It is showing you that any vector $x$ in the nullity of $B$ is also an element of the nullity of $AB$. In particular, the nullity of $AB$ must be bigger then the nullity of $B$ – Exodd Dec 8 '17 at 15:56

What is proved there is that $\operatorname{null}(B)\subset\operatorname{null}(AB)$. It follows from this that $\dim\operatorname{null}(B)\leqslant\dim\operatorname{null}(AB)$.
As a follow up from other answers, it has been shown that $$null(B) \subset null(AB)$$. Because the null space of a matrix follows the vector space axioms, the null space is a vector space. Since $$null(B) \subset null(AB)$$, $$null(B)$$ is a subspace of $$null(AB)$$.
If S is a subspace of vector space V, $$dim(S) \leq dim(V)$$. So $$nullity(B) \leq nullity(AB)$$.