Solution MM.T41 David Braithwaite

From the solution to Exercise MM.T40 we know that $\nsp{B}\subseteq\nsp{AB}$. So to establish the set equality (Definition SE) we need to show that $\nsp{AB}\subseteq\nsp{B}$.

Suppose $\vect{x}\in\nsp{AB}$. Then we know that $AB\vect{x}=\zerovector$ by Definition NSM. Consider \begin{align*} \zerovector &=\left(AB\right)\vect{x}&&\knowl{./knowls/definition.NSM.knowl}{\text{Definition NSM}}\\ &=A\left(B\vect{x}\right)&&\knowl{./knowls/theorem.MMA.knowl}{\text{Theorem MMA}} \end{align*} So, $B\vect{x}\in\nsp{A}$. Because $A$ is nonsingular, it has a trivial null space (Theorem NMTNS) and we conclude that $B\vect{x}=\zerovector$. This establishes that $\vect{x}\in\nsp{B}$, so $\nsp{AB}\subseteq\nsp{B}$ and combined with the solution to Exercise MM.T40 we have $\nsp{B}=\nsp{AB}$ when $A$ is nonsingular.