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?



Linear Algebra without Choice
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](https://mathoverflow.net/questions/93242/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?](https://mathoverflow.net/questions/80765/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?](https://mathoverflow.net/questions/49388/is-the-non-triviality-of-the-algebraic-dual-of-an-infinite-dimensional-vector-sp/79437#79437)
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?](https://mathoverflow.net/questions/49351/does-the-fact-that-this-vector-space-is-not-isomorphic-to-its-double-dual-requir/93201#93201) 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 $\sf ZF+\lnot AC$.


