Exercise 3.F.29(a)
Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V,W)$.
(a) Prove that if $\varphi \in W'$ and $\text{null} \ T' = \text{span} \ (\varphi)$, then $\text{range} \ T = \text{null} \ \varphi$.
Source.
Linear Algebra Done Right, Sheldon Axler, 4th edition.
Notation.
- $\mathcal{L}(V,W)$ is the set of all linear transformations from a vector space $V$ to $W$.
- $W' := \mathcal{L}(W,\mathbb{F})$, i.e., the dual space of $W$.
- $U^0 := \{\varphi \in W': \varphi(u) = 0 \ \text{for all} \ u \in U\}$, i.e., the annihilator of a subspace $U$ of $W$.
My attempt.
Observe that both $\text{range} \ T$ and $\text{null} \ \varphi$ are subspaces of $W$. Thus, by Exercise 21 (b) in Section 3F, we could instead show
$$ (\text{null} \ \varphi)^0 = (\text{range} \ T)^0 $$
But recall, by result 3.128, we have $(\text{range} \ T)^0 = \text{null} \ T'$. We're also given $\text{null} \ T' = \text{span} \ (\varphi)$. Thus, we will actually show
$$ (\text{null} \ \varphi)^0 = \text{span} \ (\varphi) $$
Let's first show $(\text{null} \ \varphi)^0 \subseteq \text{span} \ (\varphi)$. Let $\phi \in (\text{null} \ \varphi)^0$. Then $\phi(w) = 0$ for all $w \in \text{null} \ \varphi$. Since $\phi(w) = 0$, we have $a \phi(w) = 0$ for all $a \in \mathbb{F}$. This is where I got stuck.
My questions.
I'm tempted to just "let" or "denote" $\phi$ as $\varphi$, but I know I cannot do that in this case, right? Is it because $\varphi$ might not be the only linear functional in $W'$ such that $\varphi(w) = 0$ for all $w \in \text{null} \ \varphi$?
Am I going in the right direction? I think I'm trying to be too slick and could just follow the proof presented here: Exercise 3.F.28 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler. but I think there's a reason Axler gave the above-mentioned exercise in the 4th edition to make this proof easier.