$\operatorname{rk}(T \circ S) \leq \min\{ \operatorname{rk}T,\operatorname{rk}S \}$

Question

$T,S$ 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.

Answer 1

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.

Answer 2

For the second inequality, I would do it this way:

observe the image of $TS$ is no other than the rank of the restriction of $T$ to the image of $S$, so $$\operatorname{rk}(TS)=\dim\Bigl(\operatorname{Im}\bigl(T\big|_{\operatorname{Im}S}\bigr)\Bigr)\le\dim(\operatorname{Im}S)=\operatorname{rk}(S).$$