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.

What is proved there is that $\operatorname{null}(B)\subset\operatorname{null}(AB)$. It follows from this that $\dim\operatorname{null}(B)\leqslant\dim\operatorname{null}(AB)$.