Download 28 Some More Examples

Some More Examples
Learning Goals: take a look at a few more examples
• Subspace arithmetic
• Solution spaces for differential equations
• Infinite dimensional vector spaces
Let’s take a look at a few more examples.
Subspace arithmetic
In class challenge problem: Let V1 and V2 be subspaces of vector space W.
1) Show that any subspace of a finite-dimensional vector space is finite dimensional.
2) Show V1 ∩ V2 is a subspace
3) Let V1 + V2 = {v in W | v = v1 + v2 for some v1 in V1 and some v2 in V2} Show this is a
subspace.
4) Show, if W is finite dimensional, that dim(V1) + dim(V2) = dim(V1 ∩ V2) + dim(V1 + V2)
Function spaces
Differentiation is a linear operation (it commutes with addition and scalar multiplication).
So instead of multiplying a vector by a matrix, we can “multiply” a function by a differentiation.
So, for instance, the linear differential equation y′ = 0 is looking for the nullspace of the
derivative operator. This is, of course, the set of constant functions. Notice that this is a
subspace of the set of functions.
More generally, we could look at the linear equation a(x)y″ + b(x)y′ + c(x)y = 0. Just as
the sum of matrices is a new matrix, the sum of differential operators is a differential operator.
So we are looking at another linear operation on the set of vectors. Thus, we are looking at
another nullspace. So the solutions to this equation should be a subspace of the space of
functions.
Example: y′ + y = 0. This has solutions Ce–x for any constant C. The set of multiples of a single
vector is certainly a subspace.
Example: y″ + y = 0. This has solutions A cos(x) + B sin(x) for constants A and B. It is the set
of all linear combinations of the nullvectors cos(x) and sin(x), so is a subspace.
Example: y″ – y = 0 has the solution space Aex + Be–x.
Example: y″ – y′ = 2 does not have a subspace as its solutions, because it is not a homogeneous
equation. To solve, we find a specific solution, such as y = –2x, and add the null vectors. The
general solution is y = –2x + Aex + B.
Infinite dimensional spaces
Another interesting class of examples is infinite-dimensional vector spaces. This would
be any vector space that does not have a finite spanning set. For instance, the space P of
polynomials has no finite spanning set. How much of the stuff about bases and dimension can
we save?
It turns out we can save most of it for many vector spaces. For instance, the space P does
have a basis. One example is the infinite set {1, x, x2, x3,…}. Any polynomial is a finite linear
combination of these. So they span P. They are clearly also linearly independent, because any
nontrivial finite linear combination of the monomials will be a nonzero polynomial, so the
monomials are linearly independent. The dimension is the cardinality of this basis, which is the
smallest infinite number called ℵ0.
Let’s look at two more complicated infinite dimensional spaces. First, let’s look at the
set of convergent sequences of real numbers. Since the sum and constant multiple of convergent
sequences both converge, and all the rest of the vector space axioms can be checked componentby-component, this is a vector space. It is clearly not finite dimensional. For if you have any set
of n sequences, we can consider the first n + 1 slots. These n sequences then do not span all of
Rn+1, so we can’t even combine them to find all the different first n + 1 slots, much less all
convergent sequences.
Does it have a basis? What about a dimension?
Even weirder, let’s look at the real numbers R, but only allow the rationals as
coefficients. Now, you might think that 1 is a basis, because all reals are multiples of 1. But
they are not all rational multiples of 1! So you need to toss in a few irrational numbers. Can
you toss in enough so that you get a basis? It depends! You can get something call a Hamel
basis if you believe in the axiom of choice. But you can’t get a basis if you don’t believe in the
axiom of choice!
Note that our extension/contraction theorem doesn’t quite work, because it may take
infinitely many steps. But you are only allowed to do make infinitely many decisions if you
don’t allow the axiom of choice. So even fairly simple questions can lead to fairly complicated
answers!