How to prove that $\text{null} \;T_1 \subset \text{null} \;T_2$ implies $\text{dim(range}\; T_1) \geq \text{dim(range}\ T_2)$?

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.

• Use \ge for $\ge$ and \le for $\le$. Also, to make your own operators (like $\operatorname{dim}$ and $\operatorname{null}$), use the command \operatorname{null}. Sep 18, 2018 at 15:56
• Do you mean kernel rather than nullity, a number? Sep 18, 2018 at 15:58
• @user10354138 I'm thinking that he means "nullspace", the synonym for "kernel" that Axler uses. Sep 18, 2018 at 15:59
• In which case it follows from math.stackexchange.com/questions/2457244/… Sep 18, 2018 at 16:14

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,

$$\operatorname{dim} \operatorname{range} T_1 \ge \operatorname{dim} \operatorname{range} T_2.$$

Since you can prove it for $V$ finite-dimensional, the easiest way out is to look at the induced linear maps on the quotient $V/\ker T_1\to W$.

Let $w_1,\dots,w_r$ be a basis for $\operatorname{range}T_2$. Choose $v_1,v_2,\dots,v_r$ so that $T_2(v_i)=w_i$ for $i=1,2,\dots,r$. Finally define $u_1,\dots,u_r\in W$ by $u_i=T_1(v_i)$.

Since $(w_i)_{i=1}^r$ is a basis for $\operatorname{range}T_2$, any nontrivial linear combination of $(v_i)_{i=1}^r$ will not be in $\operatorname{null} T_2$, so it will not be in $\operatorname{null} T_1$, proving $(u_i)_{i=1}^r$ is linearly independent.