Let's say you've got a linear transformation T:V->V, and suppose V is n-dimensional. Consider exterior powers of V, Λ^k(V); for each one we naturally get a linear transformation Λ^k(T):Λ^k(V)->Λ^k(V). In particular take k=n, so we get a linear transformation Λ^n(T):Λ^n(V)->Λ^n(V). But Λ^n(V) is one-dimensional, so Λ^n(T) must be multiplication by some constant factor. That factor is the determinant.
That's basically the most "natural" definition of the determinant (notice how multiplicativity immediately falls out of it, assuming of course you already know that Λ^k is a functor). You need the idea of a basis in order to (a) make sense of the statement "V is n-dimensional" and (b) prove that Λ^n(V) is 1-dimensional, but that's it, and you certainly never need to choose one in order to define the determinant.
Probably not a great definition for beginning students of linear algebra, but I do have to correct this idea that defining the determinant requires choosing a basis...
The determinant is uniquely determined by the following two axioms:
1. The determinant of the "multiply by lambda" operation on a one-dimensional vector space is lambda. 2. If you have a linear operator T on a vector space V, and a T-invariant subspace W, the determinant of T is the product of the determinant of T restricted to W and the determinant of the operator that T induces on the quotient space V / W.
This actually kinda intuitively meshes with the volume-stretching property: if you are stretching the volume by the factor lambda_1 along one subspace and by the factor lambda_2 along some complementary subspace, clearly the overall stretch factor is lambda_1 * lambda_2.
If you aren't working over an algebraically closed field, you can just tensor with the algebraic closure of whatever field you're working with and take the determinant there. There is also a way of adapting the definition so you don't have to do this, but it makes axiom (1) a bit more complicated.
Another bonus is this definition makes the Cayley-Hamilton theorem completely trivial.
Also, you can give an analogous definition of the trace if you replace multiplication with addition, and of the characteristic polynomial if you replace lambda in axiom 1 with x - lambda.
It's a bit trickier when the dimension is infinite, but again most definitions of dimension require there to be a basis of a particular size, the difficult part is proving that this makes sense (i.e. the dimension is unique and defined).
For those who are wondering, 'every vector space has a basis' is equivalent to the Axiom of Choice.
I mean I suppose you could just restrict the definition of dimension to vector spaces that have bases, and then you wouldn't have to. I guess that's what you're implicitly suggesting, that dimension would just not apply to vector spaces that don't have bases. That would make sense. But I'm used to thinking of dimension as, well, a function of vector spaces, not a partial function, so as I was thinking of it, you have to prove they all have bases before you can use it!
What definition of _dimension_ is there that does not rely upon the existence of a base?
In the infinite case this does not trivially give you a basis as 1) the supremum could be strictly larger than the cardinality of all linearly independent sets and 2) adding an extra vector to an infinite linearly independent set doesn't increase it's cardinality, hence there is no reason for the basis to span the entire space.
You could also take the infimum of all sets that span the entire space, but you run into similar problems.
Are there any books that use this definition? I like this style of linear algebra.
(Or, if you like, you could define the tensor algebra, take a quotient of that to get the exterior algebra, and then restrict to the image of the k'th tensor power to get the k'th exterior power.)
Again, obviously you need to use bases to prove how to compute the dimension of an exterior power. But you don't need them just to define it.