1
$\begingroup$

Exercise 3.F.29(a)

Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal{L}(V,W)$.

(a) Prove that if $\varphi \in W'$ and $\text{null} \ T' = \text{span} \ (\varphi)$, then $\text{range} \ T = \text{null} \ \varphi$.


Source.

Linear Algebra Done Right, Sheldon Axler, 4th edition.


Notation.

  • $\mathcal{L}(V,W)$ is the set of all linear transformations from a vector space $V$ to $W$.
  • $W' := \mathcal{L}(W,\mathbb{F})$, i.e., the dual space of $W$.
  • $U^0 := \{\varphi \in W': \varphi(u) = 0 \ \text{for all} \ u \in U\}$, i.e., the annihilator of a subspace $U$ of $W$.

My attempt.

Observe that both $\text{range} \ T$ and $\text{null} \ \varphi$ are subspaces of $W$. Thus, by Exercise 21 (b) in Section 3F, we could instead show

$$ (\text{null} \ \varphi)^0 = (\text{range} \ T)^0 $$

But recall, by result 3.128, we have $(\text{range} \ T)^0 = \text{null} \ T'$. We're also given $\text{null} \ T' = \text{span} \ (\varphi)$. Thus, we will actually show

$$ (\text{null} \ \varphi)^0 = \text{span} \ (\varphi) $$

Let's first show $(\text{null} \ \varphi)^0 \subseteq \text{span} \ (\varphi)$. Let $\phi \in (\text{null} \ \varphi)^0$. Then $\phi(w) = 0$ for all $w \in \text{null} \ \varphi$. Since $\phi(w) = 0$, we have $a \phi(w) = 0$ for all $a \in \mathbb{F}$. This is where I got stuck.


My questions.

I'm tempted to just "let" or "denote" $\phi$ as $\varphi$, but I know I cannot do that in this case, right? Is it because $\varphi$ might not be the only linear functional in $W'$ such that $\varphi(w) = 0$ for all $w \in \text{null} \ \varphi$?

Am I going in the right direction? I think I'm trying to be too slick and could just follow the proof presented here: Exercise 3.F.28 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler. but I think there's a reason Axler gave the above-mentioned exercise in the 4th edition to make this proof easier.

$\endgroup$
5
  • $\begingroup$ This would be easier to answer if you explained the notation your are using. Like the $0$'s as exponents and that $'$ is the dual $\endgroup$
    – Digitallis
    Apr 2 at 13:50
  • $\begingroup$ @Digitallis yes, thank you for the feedback. I have edited the post. $\endgroup$
    – Paul Ash
    Apr 2 at 21:53
  • 1
    $\begingroup$ reads very well to me. Those things are always tricky. What you might use is that taking the dual ($\_^0$) twice should be the identity. Then you maybe can apply your reflex better, as the annihilator of the kernel can be very boxy to unravel. (however, you already know how to handle the other inclusion, and then you might argue with dimensions) $\endgroup$
    – Felix
    Apr 3 at 3:14
  • 1
    $\begingroup$ @Felix, I think that might just work. Technically, Axler leaves the notion of a double dual to a later exercise in the same section, but whatever. I'll use it anyway. Thanks for your help! $\endgroup$
    – Paul Ash
    Apr 3 at 21:47
  • $\begingroup$ You will have to use somewhere finally generated, an elegant way to do that is to use iso to double dual (a vector space is finitely generated if and only if the double dual is canonically isomorphic). Otherwise I fear you need to do some dimension/basis counting argument. $\endgroup$
    – Felix
    Apr 4 at 2:31

0

You must log in to answer this question.