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

dim⁡range ST≤min⁡{dim⁡range S,dim⁡range T}.\dim \text{range }ST \le \min\{\dim \text{range }S, \dim \text{range }T\}.

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

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

dim⁡range T=dim⁡range S∣R+dim⁡null S∣R≥dim⁡range S∣R\begin{aligned} \dim \text{range }T &= \dim \text{range }S|_R + \dim \text{null }S|_R \\ &\ge \dim \text{range }S|_R \\ \end{aligned}

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

dim⁡range ST≤dim⁡range T\dim \text{range }ST \le \dim \text{range }T

Proof of dim⁡range ST=dim⁡range S∣R\dim \text{range }ST = \dim \text{range }S|_R. If y∈range STy \in \text{range }ST then there exists an xx with y=S(Tx)y = S(Tx) so clearly y=S∣R(Tx)y = S|_R(Tx) and range ST⊆dim⁡range S∣R\text{range }ST \subseteq \dim \text{range }S|_R. Now suppose y∈range S∣Ry \in \text{range }S|_R, then y=Szy = Sz where z=Txz = Tx for some xx, implying y=STxy = STx and thus range S∣R⊆range ST\text{range }S|_R \subseteq \text{range }ST.

I learned two new techniques doing this exercise!