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 $\dim W'=\dim W$,
By 3.109,
Since $\varphi\in\operatorname{span}(\varphi)=\operatorname{null}T'$,
$T'(\varphi)=\varphi\circ T=0$.
So, $\operatorname{range}T\subset\operatorname{null}\varphi.$
Since $\dim\operatorname{range}T=\dim\operatorname{null}\varphi$,

  • 4
    $\begingroup$ I've noticed that every day you're posting your solutions from LADR, and asking for feedback. If you're interested, the website linearalgebras.com has solutions for most of the problems from this text, and one thing you could do is check your answers against theirs before asking for feedback here. $\endgroup$ Commented May 27, 2022 at 2:58
  • $\begingroup$ @ElliotHerrington Thank you very much for your information. $\endgroup$
    – tchappy ha
    Commented May 27, 2022 at 3:08

1 Answer 1


I just did the same problem, did the same solution, seems good to me. (Sorry for the late response).

  • 1
    $\begingroup$ Fabrizio Gambelin, Thank you very much for your answer. $\endgroup$
    – tchappy ha
    Commented Jul 27, 2022 at 10:10

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