If $W$ is finite dimensional, $T_1, T_2 \in L(V,W)$ and $\text{null}\; T_1 \subset \text{null}\; T_2$ then $$\text{dim}( \text{range}\; T_1) \geq \text{dim}(\text{range}\; T_2)$$
I'm solving a problem in Linear Algebra Done Right and I'm still struggling with a problem. And in my approach I current facing this problem. I can prove it easily when $V$ is finite dimensional but things get ugly when I attempt to prove for more general case. Any help will be appreciated.
Best Answer
There are a few ways to approach this, but here's one way: restrict $V$ to a finite-dimensional space.
Take bases $(T_1 v_1, \ldots, T_1 v_n)$ of $\operatorname{range} T_1$ and $(T_2 w_1, \ldots, T_2 w_m)$ of $\operatorname{range} T_2$. Then let $$V' = \operatorname{span} \lbrace v_1, \ldots, v_n, w_1, \ldots, w_m \rbrace \le V.$$ Note that $V'$ is finite-dimensional, and \begin{align*} \operatorname{range} (T_1|_{V'}) &= \operatorname{span} \lbrace T_1 v_1, \ldots, T_1 v_n, T_1 w_1, \ldots, T_1w_m \rbrace \\ &\supseteq \operatorname{span} \lbrace T_1 v_1, \ldots, T_1 v_n \rbrace \\ &= \operatorname{range}(T_1) \\ &\supseteq \operatorname{range}(T_1|_{V'}). \end{align*} Hence $\operatorname{range}(T_1) = \operatorname{range}(T_1|_{V'})$. Similarly, $\operatorname{range}(T_2) = \operatorname{range}(T_2|_{V'})$.
As for the nullspaces, $$\operatorname{null} T_1|_{V'} = \operatorname{null} T_1 \cap V' \subseteq \operatorname{null} T_2 \cap V' \subseteq \operatorname{null} T_2|_{V'}.$$
Since you can prove the result on finite-dimensional spaces, you now know that $$\operatorname{dim} \operatorname{range} T_1|_{V'} \ge \operatorname{dim} \operatorname{range} T_2|_{V'}.$$
But, since restriction to $V'$ didn't affect the range,
$$\operatorname{dim} \operatorname{range} T_1 \ge \operatorname{dim} \operatorname{range} T_2.$$