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.

  • $\begingroup$ Are annihilators already defined at that point? $\endgroup$ Feb 28, 2017 at 14:41
  • $\begingroup$ Yes. They are.$ $ $\endgroup$
    – Bobbie D
    Feb 28, 2017 at 14:42
  • $\begingroup$ Then it's probably better to deal with the more general case. What relation is there between $\ker T$ and $\operatorname{im} T'$? $\endgroup$ Feb 28, 2017 at 14:44
  • $\begingroup$ 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) $\endgroup$
    – Bobbie D
    Feb 28, 2017 at 14:48
  • 1
    $\begingroup$ Thanks. If you're interested, here are links to three courses that use our book. math.mit.edu/~cyxu/201618700.html and math.berkeley.edu/~gmelvin/math110f15/math110f15.html which use the 3rd ed. and math.berkeley.edu/~mhaiman/math110-spring12 which uses the 2nd ed. $\endgroup$
    – user12802
    Mar 2, 2017 at 23:46

1 Answer 1


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 $(\operatorname{im} T')^0 = \ker \varphi$.

  • $\begingroup$ Oh wow, that's a really short proof. Thanks! $\endgroup$
    – Bobbie D
    Feb 28, 2017 at 14:56

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