Linear Algebra without Choice

Question

We consider the field of "usual" linear algebra.

Q. Which aspects of it can be carried out without the Axiom of Choice?

Q. Do interesting "exotic" phenomena appear in presence of (some instance of) the negation of the Axiom of Choice?

Without Choice, vector spaces may have a basis (hence, in particular, be dimensional) or not and hence be adimensional. [As Andreas Blass observes in the comments, the terminology "dimensional/adimensional" should rather be used to denote the property of having all bases of the same cardinality, rather than just having a basis, as there are vector spaces with two bases of different cardinality]

Q. Could the following property of a vector space $V$

Property ($\star$) Every injective endomorphism of $V$ is an automorphism.

be a valid substitute for finite-dimensionality for the class of not-necessarily-dimensional vector spaces over a field? Would the linear algebra of vector spaces verifying ($\star$) be reasonably similar to the usual one for finite dimensional spaces?

Answer 1

Some things about vector spaces which are consistent with the failure of choice:

1. Vector spaces may have bases of different cardinality. In particular, this means that the notion of "dimension" is not well-defined. It follows from the Boolean Prime Ideal theorem (which is strictly weaker than $\sf AC$ itself) that if there is a basis, then its cardinality is unique. See Sizes of bases of vector spaces without the axiom of choice for more details.

2. The existence of a basis is no longer hereditary. That is, it is consistent that there is a vector space which has a basis, but it has a subspace which doesn't have a basis. You can find the example in Goldstern's answer If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?, and what is even more interesting is the fact that the vector space without a basis has a direct complement which has a well-ordered basis.

3. It is consistent that there is a vector space, that all its endomorphisms are scalar multiplications (which is not $(0)$ or the field itself). In particular every non-zero endomorphism is an automorphism, and this answers yours final question. Indeed every non-zero endomorphism is an injective endomorphism and an automorphism. These spaces were the main topic of my masters thesis, where I somewhat extended Lauchli's original result (and construction) of such spaces. You can find somewhat of an outline of the general result here: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

4. It is consistent that there is a vector space, which is not finitely generated, which is (naturally) isomorphic to its algebraic double dual. In particular this can be $\ell_2$. See my answer at Does the fact that this vector space is not isomorphic to its double-dual require choice? for details.

There are other properties which fail for non-finitely generated vector spaces in $\sf ZFC$ which are consistent with the failure of choice. The list is long, and these just a few I could write about from the top of my head.

Whether or not any of them is equivalent to the axiom of choice is usually an open (and a difficult) question. But it is usually the case that if a property requires some form of choice (like a basis, or extension of functinoals, etc.) then it can fail in suitable models of $\sf ZF+\lnot AC$.