19

I recently came across this question:

Is the axiom of choice needed to prove the following statement:

Let $V, W$ be vector spaces, and suppose $V \neq \{0\}$. Let $v \in V$, $v \neq 0$, $w \in W$. There exists a linear map $T : V \rightarrow W$ such that $Tv = w$.

I have talked to someone, who went and asked a few people, and they think that just ZF by itself is not sufficient, and that we indeed do need AC. Furthermore, they think that the statement is itself not sufficient to prove AC. Can anyone give a definitive answer?

smyrlis
  • 2,873
Sky Cao
  • 191
  • 4
  • 5
    It is consistent with ZF that there is a vector space whose dual space is zero, so you indeed need some form of choice. See http://mathoverflow.net/questions/49388/is-the-non-triviality-of-the-algebraic-dual-of-an-infinite-dimensional-vector-sp – Mariano Suárez-Álvarez Mar 24 '14 at 09:09

1 Answers1

18

Yes, the axiom of choice is needed.

Läuchli has constructed a vector space whose only endomorphisms are scalar multiplication. In such vector space, if $v\neq \alpha w$ there is no such $T$. The paper is in German, and it uses mathematical terminology which took some time to understand.

Läuchli, H. "Auswahlaxiom in der Algebra", Comment. Math. Helv. 37 (1962-1963), 1–18.

In my masters thesis I showed (by extending Läuchli's argument) that in fact $\sf ZF+DC_\kappa$ cannot prove the existence of such $T$, for any given $\kappa$. You can find a summary of the result (in a very early edition) here: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

To my knowledge, the question whether or not the assertion that such $T$ always exists implies the axiom of choice is still open.

Community
  • 1
Asaf Karagila
  • 38,140