若 \(W'\subseteq W\) 為子空間，定義限制映射
\[
F':W'\to U,\qquad F'(w)=F(w)\quad (\forall w\in W').
\]

欲證：
\[
F'(W')=F(W'),\qquad N(F')=N(F)\cap W'.
\]

證明 \(F'(W')=F(W')\)：

由定義可知，對任意 \(w\in W'\)，
\[
F'(w)=F(w).
\]
因此
\[
F'(W')=\{F'(w):w\in W'\}=\{F(w):w\in W'\}=F(W').
\]

證明 \(N(F')=N(F)\cap W'\)：

由 kernel 的定義，
\[
N(F')=\{w\in W':F'(w)=0\}.
\]
但因為 \(F'(w)=F(w)\)，故
\[
N(F')=\{w\in W':F(w)=0\}
= \{w\in W':w\in N(F)\}
= N(F)\cap W'.
\]

因此已證
\[
F'(W')=F(W'),\qquad N(F')=N(F)\cap W'.
\]

接著由 rank-nullity theorem 套用於 \(F':W'\to U\)，得
\[
\dim(W')=\dim(N(F'))+\dim(F'(W')).
\]
利用上面已證結果，
\[
\dim(W')=\dim(N(F)\cap W')+\dim(F(W')).
\]
故
\[
\dim(N(F)\cap W')=\dim(W')-\dim(F(W')).
\]
又因為
\[
N(F)\cap W'\subseteq N(F),
\]
所以
\[
\dim(N(F))\ge \dim(N(F)\cap W')
=\dim(W')-\dim(F(W')).
\]
故證得
\[
\boxed{\dim(N(F))\ge \dim(W')-\dim(F(W')).}
\]

令 \(W'=T(V)\)。

由(1)可得
\[
\dim(N(F))\ge \dim(T(V))-\dim(F(T(V))).
\]
注意
\[
\dim(T(V))=\operatorname{rank}(T),\qquad F(T(V))=(F\circ T)(V),
\]
所以
\[
\dim(F(T(V)))=\operatorname{rank}(F\circ T).
\]
故
\[
\dim(N(F))\ge \operatorname{rank}(T)-\operatorname{rank}(F\circ T).
\]
再由
\[
\operatorname{rank}(F)=\dim(W)-\dim(N(F)),
\]
得
\[
\operatorname{rank}(F)
\le \dim(W)-\bigl(\operatorname{rank}(T)-\operatorname{rank}(F\circ T)\bigr).
\]
整理得
\[
\boxed{\operatorname{rank}(F)\le \dim(W)-\operatorname{rank}(T)+\operatorname{rank}(F\circ T).}
\]

證明
\[
\operatorname{rank}(F)+\operatorname{rank}(T)-\dim(W)
\le \operatorname{rank}(F\circ T)
\le \min\{\operatorname{rank}(T),\operatorname{rank}(F)\}.
\]

\underline{左邊不等式：}

由(2)
\[
\operatorname{rank}(F)\le \dim(W)-\operatorname{rank}(T)+\operatorname{rank}(F\circ T),
\]
移項得
\[
\boxed{\operatorname{rank}(F)+\operatorname{rank}(T)-\dim(W)
\le \operatorname{rank}(F\circ T).}
\]

\underline{右邊不等式：}

因為
\[
(F\circ T)(V)=F(T(V)),
\]
所以其像空間是 \(F\) 在 \(T(V)\) 上的像，因此
\[
\operatorname{rank}(F\circ T)=\dim(F(T(V)))\le \dim(T(V))=\operatorname{rank}(T).
\]
另一方面，
\[
F(T(V))\subseteq F(W),
\]
所以
\[
\operatorname{rank}(F\circ T)=\dim(F(T(V)))\le \dim(F(W))=\operatorname{rank}(F).
\]
因此
\[
\boxed{\operatorname{rank}(F\circ T)\le \min\{\operatorname{rank}(T),\operatorname{rank}(F)\}.}
\]

綜合可得
\[
\boxed{
\operatorname{rank}(F)+\operatorname{rank}(T)-\dim(W)
\le \operatorname{rank}(F\circ T)
\le \min\{\operatorname{rank}(T),\operatorname{rank}(F)\}.}
\]