Suppose $U$ and $V$ are finite-dimensional vector spaces and $S \in \mathcal L(V,W)$ and $T \in \mathcal L(U,V)$. Prove that

Since $\text{range }ST \subseteq \text{range }S$ we immediately have $\dim \text{range }ST \le \text{range }S$
meaning it suffices to show $\dim \text{range }ST \le \text{range }T$.

Consider $S|_R$ as a map from $\text{range }T$ to $\text{range }S$. Apply 3.22 to get

It suffices to show $\dim \text{range }ST = \dim \text{range }S|_R$ since it would imply

Proof of $\dim \text{range }ST = \dim \text{range }S|_R$. If $y \in \text{range }ST$ then there exists an $x$ with $y = S(Tx)$ so clearly $y = S|_R(Tx)$ and $\text{range }ST \subseteq \dim \text{range }S|_R$. Now suppose $y \in \text{range }S|_R$, then $y = Sz$ where $z = Tx$ for some $x$, implying $y = STx$ and thus $\text{range }S|_R \subseteq \text{range }ST$.

I learned two new techniques doing this exercise!

• Restricting linear maps then applying 3.22 to get inequalities
• Compositions $ST$ can be analyzed by conditioning on the domain. ie. $\text{range }ST = \text{range }S|_{\text{range }T}$ and $\dim \text{null }ST = \dim \text{null }S|_{\text{range }T} + \dim \text{null }T$