4
$\begingroup$

A problem from Linear Algebra Done Right (Third Ed): Suppose $W$ is finite dimensional and $T \in \mathcal{L}(V, W)$. Prove that $T=0$ if and only if its dual $T' \in \mathcal{L}(W', V')=0$.

I am confused by the information that $W$ is finite dimensional. Why is that required, while nothing is said about finite dimensionality of $V$ ?

There is a solution here: Does that fact that the dual map is zero imply that the map is zero? but I am more interested in the reason for the assumptions.

I was able to solve it (or so I thought) without using a contrapositive argument or the assumption on $W$ which is what worries me.

EDIT: Here is my argument. Since T' = 0, we get $\phi_w(Tv) = 0 \forall \phi_w$, which yields Tv = 0 for all v, hence T = 0. I can "reverse" the argument to get the converse. What is my error?

$\endgroup$
1
  • 2
    $\begingroup$ Maybe you should add your argument. $\endgroup$ Commented Feb 14, 2016 at 10:50

1 Answer 1

Reset to default
5
$\begingroup$

Note that the statement is correct also for infinite-dimensional $W$, as the key point is the

Proposition. If $W$ is a vector space, $w \in W$ such that $f(w)=0$ for all $f \in W'$, then $w = 0$.

(This follows from the fact that if $w \ne 0$, we can extend $\{w\}$ to a basis $B$ of $W$ and define $f$ on the basis $B$ by $f(w) = 1$ and $f(x) = 0$ for $x \in B \setminus \{w\}$, and extend by linearity. Then $f \in W'$ and $f(w) \ne 0$).

But this argument relies on the axiom of choice (this is always needed if we want to construct bases on infinite-dimensional vector spaces), and may be wrong if the axiom of choice is not to be used. As far as I know, in the mentioned book, the basis extension theorem is not proved for infinite-dimensional vector spaces, and hence cannot be used.

As we do not need a basis of $V$ in our argument, not dimensionality assumption if needed for $V$.

$\endgroup$
5
  • $\begingroup$ See here on mathoverflow for an example of a infinite-dimensional vector space $V$ such that $V' = 0$. $\endgroup$ Commented Feb 14, 2016 at 14:47
  • 3
    $\begingroup$ The explanation above by martini is correct. In my book Linear Algebra Done Right, the focus is on finite-dimensional vector spaces. Thus the book does not prove that every infinite-dimensional vector space has a basis (the proof requires the use of the axiom of choice or an equivalent result). Hence this exercise has the hypothesis that W is finite-dimensional. The solution to the exercise does not need to use a basis of V, and thus I did not include a finite-dimensionality hypothesis for V in the exercise. $\endgroup$ Commented Feb 15, 2016 at 2:24
  • $\begingroup$ This chain of equivalences seems to work for the infinite-dimensional cases without invoking the axiom of choice: \begin{equation} T=0\iff\mathrm{range}\,T=\{0\}\iff(\mathrm{range}\,T)^0=W'=\mathrm{null}\,T'\iff T'=0. \end{equation} Can you see any flaw in it? Thanks. $\endgroup$ Commented Jun 14, 2021 at 2:09
  • 2
    $\begingroup$ The chain of equivalences above does use the axiom of choice in the case when $W$ is infinite-dimensional. Without the axiom of choice, $W'$ might equal $\{0\}$, and thus the backward direction of your second if and only if might not hold. $\endgroup$ Commented Jun 14, 2021 at 5:26
  • $\begingroup$ Thanks for the answer. I understand that $W'=\{0\}$ breaks the second iff. But I didn't get why with the Axiom of Choice, $W'$ might not equal $\{0\}$? $\endgroup$ Commented Jun 14, 2021 at 17:45

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.