$\operatorname{rk}(T \circ S) \leq \min\{ \operatorname{rk}T,\operatorname{rk}S \}$ $T,S$ are both linear transformations.
How do I approach this proof? it is rather intuitive why this "restriction" takes place, although I find it kinda hard proving it.

Step 1: $\;
\operatorname{rank}(TS)\leq \operatorname{rank}(T)$
$S(V)\subset V$ which implies that $TS(V)\subset T(V)$ therefore rank of $TS$ is less than or equal to $T$
Step 2: $\;\operatorname{rank}(ST)\leq \operatorname{rank}(T)$
Suppose $\operatorname{rank}(T)=m$. It mean $T(V)$ has dimension $m$. Let us denote the basis vectors by $w_1, w_2, \dots w_m$. Consider $Sw_1, \dots Sw_m$. Conclude that dimension of $ST$ can be at most $m$.
Use Step 1 and Step 2 to conclude the required result.
Reference: Lemma 6.13 I N Herstein, Topics in Algebra, Second Edition.
