I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
I solved Exercise 3.F.28.
This exercise is very abstract for me.
So, I am not sure that my solution is ok or not.
3.94 Definition dual space, $V'$
The dual space of $V$, denoted $V'$, is the vector space of all linear functionals on $V$. In other words, $V'=\mathcal{L}(V,\mathbb{F})$.
3.99 Definition dual map, $T'$
If $T\in\mathcal{L}(V,W)$, then the dual map of $T$ is the linear map $T'\in\mathcal{L}(W',V')$ defined by $T'(\varphi)=\varphi\circ T$ for $\varphi\in W'.$
Exercise 3.F.28
Suppose $V$ and $W$ are finite-dimensional, $T\in\mathcal{L}(V,W)$, and there exists $\varphi\in W'$ such that $\operatorname{null}T'=\operatorname{span}(\varphi).$ Prove that $\operatorname{range}T=\operatorname{null}\varphi.$
I used the following proposition on p.107 in this book:
3.109 The range of $T'$
Suppose $V$ and $W$ are finite-dimensional and $T\in\mathcal{L}(V,W).$ Then
(a) $\dim\operatorname{range}T'=\dim\operatorname{range}T;$
My solution:
Assume that $\varphi=0$.
Then, $\operatorname{null}\varphi=W.$
Then, $\operatorname{null}T'=\operatorname{span}(\varphi)=\{0\}$.
By Fundamental Theorem of Linear Maps (3.22 on p.63),
$\dim W'=\dim\operatorname{null}T'+\dim\operatorname{range}T'=\dim\operatorname{range}T'.$
Since $\dim W'=\dim W$ and $\dim\operatorname{range}T'=\dim\operatorname{range}T$ by 3.109,
$\dim W=\dim\operatorname{range}T.$
So, $\operatorname{range}T=W=\operatorname{null}\varphi.$
Assume that $\varphi\neq 0$.
Then, $\dim\operatorname{null}T'=\dim\operatorname{span}(\varphi)=1.$
By Fundamental Theorem of Linear Maps (3.22 on p.63), $\dim W'=\dim\operatorname{null}T'+\dim\operatorname{range}T'=1+\dim\operatorname{range}T'.$
Since $\varphi\neq 0$, $\dim\operatorname{range}\varphi=1.$
By Fundamental Theorem of Linear Maps (3.22 on p.63),
$\dim W=\dim\operatorname{null}\varphi+\dim\operatorname{range}\varphi=\dim\operatorname{null}\varphi+1.$
Since $\dim W'=\dim W$,
$\dim\operatorname{range}T'=\dim\operatorname{null}\varphi.$
By 3.109,
$\dim\operatorname{range}T=\dim\operatorname{null}\varphi.$
Since $\varphi\in\operatorname{span}(\varphi)=\operatorname{null}T'$,
$T'(\varphi)=\varphi\circ T=0$.
So, $\operatorname{range}T\subset\operatorname{null}\varphi.$
Since $\dim\operatorname{range}T=\dim\operatorname{null}\varphi$,
$\operatorname{range}T=\operatorname{null}\varphi.$
