This lecture is intended as a foretaste of things to come. We will develop extra layers of abstraction, and this will allow us to apply the ideas of linear algebra in new contexts.
They key operations we've used on vectors are additon and rescaling: v + w, and lambda times v. These operations are all we need to state the definitions of "linear map" and "linear subspace". But we can add and rescale things much more general than column vectors in R^n.
Definition:
A (real) vector space is a set V with operations add from V times V to V written add of v, w = v + w, and rescale from R times V to V written rescale of lambda, v = lambda times v. The elements of V don't have to be elements of R^n, and add and rescale don't have to be addition and rescaling of vectors in R^n, they could be any other objects and operations which behave in a similar manner, by which I mean that the following conditions hold (for all u, v, w in V and lambda in R):
There is a distinguished element 0 in V.
u + the sum (v + w) = the sum (u + v) plus w, i.e. associativity of addition
v + w = w + v
v = 0 + v = v + 0
v + minus v = 0
1 v = v
lambda times (mu times v) = lambda mu times v
lambda + mu all times v = lambda v + mu v
lambda times the sum v + w equals lambda v + lambda w.
So a vector space is a set with operations that satisfy these conditions. In particular, R^n equipped with vector addition and rescaling is an example of a vector space. Indeed, it is a theorem that any finite-dimensional vector space is equivalent to R^n for some n. But if you allow yourself to consider infinite-dimensional vector spaces there are more interesting examples.
Example:
Let V be the set of all functions from R to R. Given functions f, g, define f + g to be the function whose value at x in R is f of x + g of x, and given a scalar lambda in R we define lambda f to be the function whose value at x in R is lambda times f of x. This gives V the structure of a vector space. It's not the same as R^n for any n: it is actually an infinite-dimensional vector space.
Example:
Inside our previous example we can find some linear subspaces. For example, the set of continuous functions C^0(R) is a subspace. Inside that we have another subspace: the space of once continuously-differentiable functions C^1(R). Inside that we have the subspace of twice continuously-differentiable functions C^2(R). And so on. Inside all of these we have the infinitely-differentiable functions C^{infinity}(R), and, inside that, that space of analytic functions C^{omega}(R) (infinitely-differentiable functions whose Taylor series converges on a neighbourhood of the origin). This gives us an infinite nested sequence of subspaces: C^{omega}(R) inside C^{infinity}(R) inside dot dot dot inside C^2(R) inside C^1(R) inside C^0(R) Inside the space of analytic functions, we have the space of polynomials (which we already met).
Example:
Differentiation defines a linear map d by d x from C^1(R) to C^0(R), sending f to d f by d x In other words, you start with a once continuously-differentiable function and differentiate it to get a continuous function). To show that it's linear, all we need to do is check that d by d x of f + g equals d f by d x + d g by d x, and d by d x of lambda f = lambda times d f by d x.
What is the kernel of d by d x? It consists of functions whose derivative is zero, in other words constant functions. That is, ker of d by d x inside C^1(R) is the 1-dimensional subspace of constant functions.
The eigenfunctions of d by d x (with eigenvalue lambda) will be a function f such that d f by d x = lambda f. This is a differential equation for f; its solution is f of x equals C times e to the lambda x. This is why the exponential functions is so important: it's an eigenfunction of differentiation. Similarly the eigenfunctions of d 2 by d x squared are the solutions of d 2 f by d x squared = lambda f, that is A cos x root minus lambda + B sine x root minus lambda. This is a 2-dimensional eigenspace.
Algebraic numbers
Example:
Here is an example from number theory. We have come across the idea that your vectors can have complex coefficients or real coefficients, but we can work much more generally by requiring our coefficients to live in some coefficient field K. In this example, we will take K = Q (the rational numbers), but you could imagine all sorts of things (the integers modulo 5, the 17-adic numbers, and goodness knows what else). The only difference this makes in the definition of a vector space is that rescaling can only be done by elements of K, that is the rescaling map is rescale from K times V to V.
A number x in C is algebraic if there exists a polynomial P of z = a_n z to the n + dot dot dot + a_0 with a_n, a_{n minus 1}, dot dot dot, a_0 in Q for which P of x = 0.
For example, root 2 is an algebraic number because it's a root of z squared = 2.
i is algebraic because it's a root of z squared = minus 1.
pi and e are not algebraic (i.e. they're transcendental)
The set of all algebraic numbers is called Q bar.
Lemma:
Q bar is a Q-vector space.
We need to show that if x, y are algebraic numbers and lambda is a rational number then x + y and lambda x are algebraic numbers.
To see that lambda x is in Q bar, note that there is a polynomial P of z = a_n z to the n + dot dot dot + a_0 with all a_k in Q and P of x = 0. Now R of z = a_n times z to the n over lambda to the n + dot dot dot + a_1 z over lambda + a_0 satisfies R of lambda z = P of z, so R of lambda x = P of x = 0, showing that lambda x is in Q bar. Note that we really need lambda to be in Q (or else the coefficients a_k over lambda to the k are not rational), so this is only a Q-vector space (not a R-vector space).
To show that x + y is algebraic is much harder: we can't prove it here. In the words of Pierre Samuel (Algebraic Theory of Numbers), "The reader will have to exert himself to show that the cube root of 5 plus the cube root of 7 is an algebraic integer, and will be convinced that the steps which lead to a proof that this number is algebraic may not be easily generalised." The nicest proof uses the theory of modules over rings.
In fact, slightly more is true: the product of two algebraic numbers is also an algebraic number, so Q bar is a field. One of the most complicated objects in mathematics is the group Gal Q bar over Q, the Galois group of \mathbf{Q}. This is the set of invertible Q-linear maps f from Q bar to Q bar which preserve the product (f of a b equals f of a times f of b).
The elements of Gal Q bar over Q are like infinitely large matrices with rational entries (because Q bar is infinite-dimensional over Q). One way people study this enormously complicated group is using Galois representations: associating to each g in the Galois group a finite-dimensional matrix M of g such that M of g_1 times g_2 = M of g_1 times M of g_2. Constructing Galois representations is a very difficult task, but the payoff can be enormous. Galois representations played an important role in Wiles's proof of Fermat's last theorem, and continue to play an important role in modern number theory.