I've already proven this question, however, my proof is much more complicated than an answer I saw posted on another website. However, I cannot seem to make sense of why the answer on the website is correct.$\DeclareMathOperator{\range}{range}\DeclareMathOperator{\null}{null}\DeclareMathOperator{\span}{span}$
The answer on the website uses the key result that $\null(T') = (\range(T))^0$ to conclude:
If $\null(T') = \span(\phi)$, then using the result above, we get $(\range(T))^0 = \span(\phi)$
This implies that $w \in \range(T)$ if and only if $\phi(w) = 0$.
I see how the forward direction of the above “if and only if” is true, in fact, I used it in my proof. However, how is it obvious that $\phi(w) = 0 \implies w \in \range(T)$? In light of my later proof, I see that it is true, but I don't see how we can conclude this right off the bat.
To give you an idea of what I did, my proof first showed that $\range(T) \subseteq \null\phi$ (using the forward direction of bolded “if and only if”), and then used the three facts
$\dim \null(\phi)) = \dim W - 1$
$(\range(T))^0 = \null(T')$
$\dim(U^0) + \dim(U) = \dim(W)$ (where $U$ is any subspace)
to deduce that $\dim(\range(T)) = \dim(W) - \dim(\range(T))^0 \\ = \dim(W) - \dim(\null(T')) \\ = \dim(W) - 1$
The last equality follows from $\null(T') = \span(\phi)$, which is one-dimensional. I then concluded that since $\dim(\range(T)) = \dim(\null(\phi))$, and $\range(T) \subseteq \null(\phi)$, they must be equal.
Notation:
$W'$ is the dual space of $W$
$T'$ is the dual map of $T$
$U^0$ is the annihilator of $U$