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From Linear Algebra Done Right, Third Edition, 3.F Exercise 29:

Suppose $V$ and $W$ are finite-dimensional, $T\in \mathcal L(V,W)$, and there exists $\varphi \in V'$ such that $\operatorname{range}T' = \operatorname{span}(\varphi)$. Prove that $\operatorname{null} T = \operatorname{null} \varphi$.

It seems like every time I've got to prove two sets are equal, one inclusion is do-able but I get stuck on the other one. Oh well, here's what I have for this one:

First we show that $\operatorname{null}T\subseteq \operatorname{null}\varphi$. Let $v\in \operatorname{null} T$. Then $T(v) = 0$.
Case 1: If $\varphi = 0$, then $\operatorname{null}\varphi = V$ and hence trivially $\operatorname{null}T\subseteq \operatorname{null}\varphi$.
Case 2: If $\varphi \ne 0$, then there exists an $\alpha \in W'$ such that $T'(\alpha) = \alpha \circ T \ne 0$. Then $$0 = \alpha(0) = \alpha(T(v)) = (\alpha \circ T)(v) = (T'(\alpha))(v) = k\varphi(v)$$ for some $k\in \Bbb F$. Because $\alpha \circ T = k\varphi \ne 0$, we conclude $k\ne 0$ and thus $k\varphi (v) = 0$ implies $\varphi(v) = 0$. So $v\in \operatorname{null}T$ implies $v\in\operatorname{null}\varphi$. Hence $\operatorname{null}T \subseteq \operatorname{null}\varphi$.

Now we complete the proof by either showing that $\operatorname{null}\varphi \subseteq \operatorname{null}T$ or $\dim \operatorname{null}T = \dim \operatorname{null}\varphi$.
Case 1: If $\varphi = 0$, then $\operatorname{null}\varphi = V$ but then also $\operatorname{range}T' = \{0\}$. By definition then, $T' = 0$, and by [a previous exercise] $T=0$. Thus $\operatorname{null} T = V = \operatorname{null}\varphi$.
Case 2: If $\varphi \ne 0$, then ... $\dim\operatorname{range}T' = 1$ is all I can come up with. This is where I'm stuck.

Note: right before this was a very similar exercise which I was eventually able to complete: 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 didn't see any way to use this previous lemma to prove the present one, but if there is some clever way of showing they're equivalent, that might be an easier way to prove it.

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    Are annihilators already defined at that point? – 2017-02-28
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    Yes. They are.$ $ – 2017-02-28
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    Then it's probably better to deal with the more general case. What relation is there between $\ker T$ and $\operatorname{im} T'$? – 2017-02-28
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    Is that you just suggesting a rewrite of the question or do you want me to answer? Because I don't know. $\ker T \subseteq V$ while $\operatorname{im} T' \subseteq V'$. Are they dual spaces? If so, I don't think I've proven that yet. Edit: Nevermind $\operatorname{im} T' = (\ker T)^0$. OK. Thanks for the hint. I'll see if I can use that. :) (I don't know HOW I'm going to remember all these lemmas o.O) – 2017-02-28
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    Well, we need it in the form $\ker T = (\operatorname{im} T')^0$. Or perhaps $\ker T = {}^0(\operatorname{im} T')$, depends on the used notation. – 2017-02-28
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    Hi - I'm a self-studier working on this text. May I ask what class you are doing this for as it's always nice to see some related material and especially what problems (out of 500+) are being emphasized. Thanks. With regards, – 2017-03-02
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    @TheBirdistheWord I'm also a self-learner. I've actually already taken a class in linear algebra, but it wasn't very good (we barely covered anything), so I picked up this book and started working through it on my own. – 2017-03-02
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    Thanks. If you're interested, here are links to three courses that use our book. http://math.mit.edu/~cyxu/201618700.html and https://math.berkeley.edu/~gmelvin/math110f15/math110f15.html which use the 3rd ed. and https://math.berkeley.edu/~mhaiman/math110-spring12/ which uses the 2nd ed. – 2017-03-02

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In the following, every line is "obviously" (obvious for an experienced reader, but it should be easy enough to see also for the learner) equivalent to the one immediately following/preceding it:

\begin{gather} v \in \ker T \\ T(v) = 0 \\ \bigl(\forall \alpha \in W'\bigr)\bigl(\alpha(T(v)) = 0\bigr) \\ \bigl(\forall \alpha \in W'\bigr)\bigl((T'\alpha)(v) = 0\bigr) \\ \bigl(\forall \psi \in \operatorname{im} T'\bigr)\bigl(\psi(v) = 0\bigr) \\ v \in (\operatorname{im} T')^0 \end{gather}

Now if $\operatorname{im} T' = \operatorname{span} \varphi$, then $(\operatorname{im} T')^0 = \ker \varphi$.

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    Oh wow, that's a really short proof. Thanks! – 2017-02-28