## Transcript

It is rare to show that something is a vector space using the defining properties. Instead, most things we want to study actually turn out to be a subspace of something we already know to be a vector space.

Suppose that V is a vector space and that U is a subset of V. If U is a vector space using the same definition of addition and scalar multiplication as V, then U is called a subspace of V.

For example, is P2 a subspace of P3? Yes. Since every polynomial of degree up to 2 is also a polynomial of degree up to 3, P2 is a subset of P3. And we already know that P2 is a vector space, so it is a subspace of P3. However, R2 is not a subspace of R3 since the elements of R2 have exactly two entries, while the elements of R3 have exactly three entries. That is to say, R2 is not a subset of R3. Similarly, M(2, 2) is not a subspace of M(2, 3) because M(2, 2) is not a subset of M(2, 3).

So why are subspaces important? Well, it turns out that a subspace inherits most of the vector space axioms from its parent vector space. For example, if x + y = y + x in V, then this will continue to be true in U. That’s because this property is actually a property of the definition of addition, not of V or of U, and since our definition of a subspace requires that U use the same definition of addition, we see that addition in U must also be commutative.

Similarly, we immediately get that U satisfies any properties that only use the definition of addition and scalar multiplication. These properties are V2, that addition is associative; V5, that addition is commutative; V7, that scalar multiplication is associative; V8, that scalar addition is distributive; V9, that scalar multiplication is distributive; and V10, the scalar multiplicative identity. In fact, given any subset (but not necessarily a vector space) W of a vector space V, we know that properties V2, V5, V7, V8, V9, and V10 will hold in W.

So, if we want to prove that W is itself a vector space, we only need to look at properties V1, V3, V4, and V6. Now, properties V1 and V6 were trivial when showing that Rn or M(m, n) and Pn were vector spaces, but this property becomes much more important when we are looking at subspaces.

Consider the following. Let V be M(2, 2) and let W be the subset of M(2, 2) consisting of matrices with at most one non-zero entry. So for example, the matrix A, which is [1, 0; 0,
0]; the matrix B, [0, 5; 0, 0]; and C, [0, 0; 0, 0] are all elements of W. But let’s look at the sum A + B. We know that A + B is in V, and even that A + B = B + A, since A and B are elements of V. But it turns out that A + B is not in W since, of course, the sum, the matrix A + B is the matrix [1, 5; 0, 0], and thus, A + B has more than one non-zero entry.

And so, when we are considering properties V1 and V6, we are not so much concerned with the existence of x + y and sx as we are that these elements are contained in our smaller set. Similarly, our concern with V3 and V4 is not that 0 and inverse-x exist and satisfy their respective properties, but rather that they are actually elements of W. But it turns out that this key fact follows from property V6, closure under scalar multiplication.

Why is that? Well, let x be any element of W, and suppose that we have already shown that W is closed under scalar multiplication. Then we know that sx is in W for any s in
R, but this means that sx is in W for s = 0 and s = -1, and by Theorem 4.2.1, we know that 0x = 0 and (-1)x = inverse-x. Again, we’re now using the fact that we already know V is a vector space. And so we see that 0 is in W and inverse-x is in W, as desired.

There is one fine detail I skipped over, though, and that is the fact that I assumed that x is an element of W. Well, what’s wrong with that, you wonder. Well, it only works if W actually contains elements. That is, we cannot have that W equals the empty set. Recall that the empty set can never be a vector space since any vector space must contain at least a 0 vector.

I’ll summarize these results with the following alternate definition of a subspace. Suppose that V is a vector space. Then U is a subspace of V if it satisfies the following three properties:

- S0) U is a non-empty subset of V.
- S1) x + y is in U for all x and y in U.
- S2) tx is in U for all x in U and t in R.

Let’s look at an example. Show that the set U, which is the set of all (a + bx + c(x-squared)) in P2 where a = b and also equal to c is a subspace of P2. First, we’ll check property S0. Well, first we see that U is specifically defined as a subset of P2, and next, we’ll notice that the 0 polynomial is in U since 0 = 0 = 0, so U is not the empty set. Next, we check S1. Suppose that P and Q are elements of U. Well, then there are real numbers p and q such that p(x) = p + px + p(x-squared) and q(x) = q + qx + q(x-squared). And then we have that p(x) + q(x) will equal this sum, and if we add our coefficients, we get that it equals the polynomial (p + q) + (p + q)x + (p + q)(x-squared), and this polynomial satisfies the defining properties of U. So we’ve seen that p + q is an element of U. Finally, we check property S2. Suppose that p is in U and that s is in R. Then we’ll let our real number p be such that p(x) = p + px + p(x-squared), and we have that s(p(x)) is, of course, the polynomial sp + (sp)x + (sp)(x-squared), and since this satisfies the defining properties of U, we see that s(p(x)) is in U. And since U satisfies our properties S0, S1, and S2, we have that U is a subspace of P2.

Let’s look at another example. Let’s show that the set A = {a + bx in P1 such that b = a-squared} is not a subspace of P1. Now, to show that something is not a subspace, we need to show that any one of the three properties does not hold. Of course, figuring out which property fails can be tricky. S0 is so easy to check that we usually start there. In this case, A is obviously a subset of P1, and we quickly see that the 0 polynomial is an element of A, so it is non-empty.

So S0 is not the property that fails to hold, but what about S1? Well, if we had two functions in A, say a + (a-squared)x and b + (b-squared)x, then we would add them to get (a + b) + (a-squared + b-squared)x. Now for this to be a function in A, we need to have that a-squared + b-squared = quantity (a + b)-squared. This is, of course, not true in general. It is sometimes true, however, so instead of using a blanket statement of a-squared + b-squared is not equal to quantity (a + b)-squared to show that S1 fails, the correct course of action is to find specific values of a and b such that a-squared + b-squared is not equal to the quantity (a + b)-squared, and use them as our counterexample. One possible choice is a = 1 and b = 2, and using these values, I get the following proof that A is not a subspace of P1. A is not a subspace of P1 since there are elements 1 + x and 2 + 4x in A such that (1 + x) + (2 + 4x) = 3 + 5x, which is not in A, so A is not closed under addition. There we are, short and simple.

But before we move on, I want to make a couple of comments about this example. First, I want to remind you that any time you are trying to prove that something is not a subspace because properties S1 or S2 fail, you should use specific counterexamples. Many students want to stop with the fact that a-squared + b-squared would not equal to (a + b)-quantity-squared, but as I pointed out, this statement is sometimes true, so how do we know that is isn’t true for the specific elements of A? This particular statement would be true whenever a or b = 0, so at a minimum, we already need to verify that there are two non-zero elements of, in A, which, of course, is exactly what I did in my counterexample. Basically, continuing on a theoretical path is likely to be more work than just using your information to generate the needed counterexample. So, while noticing that the statement a-squared + b-squared = (a + b)-quantity-squared is not true was a key step in finding my counterexample, the statement does not ever need to actually appear in my proof. Use it and lose it, I say. And yes, this means that you do not need to show any work for how you find your counterexample.

Now, the last thing I want to point out is that S2 also fails to hold in our example. One counterexample could be that 2 + 4x is in A, but 5(2 + 4x), which equals 10 + 20x, is not in A. Again, we only need to show that one of the three properties fails, but since this is an example, I thought I would show this alternate possibility.

Now, remember that the point of subspaces is that they are a quick way to show that something is a vector space. For example, the set D of differentiable functions over R is a vector space. We see this by considering it as a subspace of F, the space of all functions. Then we only need to check the three subspace properties. S0: Differentiable functions are, of course, functions, so D is a subset of F. Moreover, D is not empty since, for example, the 0 function is differentiable. S1: Suppose that f and g are differentiable functions. Then f + g is also differentiable. In fact, the derivative of (f + g) is (the derivative of f) + (the derivative of g). So f + g is in D. And similarly, we’ll look at S2, and suppose that f is a differentiable function, and that s is in R. Then sf is also differentiable. In fact, the derivative of (sf) = s(the derivative of f), so sf is in D.

This same technique tells us that the set C of continuous functions is a vector space, as is the set C(a, b) of continuous functions on the interval (a, b). Even our polynomial spaces can be thought of as subspaces of F.