Transcript — Introduction
In the last lecture, we introduced the concept of an abstract vector space. We saw that this was defined to be a set with two operations of addition and scalar multiplication that satisfied the same ten properties as vectors in Rn. In this lecture, we will look at several more examples of determining if a set with two operations is a vector space or not. We will then look at spaces which are subsets of other spaces.
We begin with some more examples. Is the empty set a vector space? The fact that we can’t define meaningful operations of addition and scalar multiplication on a set with no elements worries us, since the operations are key parts of the definition of a vector space. Hence, we highly suspect that the empty set is not a vector space. Indeed, property V4 says that a vector space must contain at least one vector—the zero vector. Therefore, the empty set cannot be a vector space.
Example: Let V be the set of points (x, y) such that x and y are real. Define addition in this set by (x1, y1) + (x2, y2) = (2x1 + x2, y1 + 2y2), and define scalar multiplication by t(x1, y1) = (tx1, ty1) for any real scalar t. Is V a vector space under these operations? To prove that V is a vector space, we would need to show that all ten properties V1 to V10 hold. On the other hand, to prove that V is not a vector space, we would just need to find one example where one of the properties does not hold. In general, it is worthwhile to spend a little time thinking about the definitions of addition and scalar multiplication before deciding whether you should be trying to prove all ten properties, or if you should be trying to look for a counterexample to one of the properties. It can waste a lot of time if you are trying to prove properties of a set that is not a vector space, or if you are trying to find a counter-example which doesn’t exist since the set is a vector space. So, let’s think about the operations in this example. First, we notice that the definition of addition is a little odd, but the definition of scalar multiplication is the standard one. Therefore, if there is a problem, it will be with addition. What could be wrong with addition? Upon inspection, we see that the first and second vector are treated slightly differently in the definition of addition. This should make us think that addition should not satisfy property V3. Indeed, if we take the points (1, 2) and (3, 5), we get that (1, 2) + (3, 5) = (2(1) + 3, 2 + 2(5)) = (5, 12), but (3, 5) + (1, 2) = (2(3) + 1, 5 + 2(2)) = (7, 9). Therefore, V3 does not always hold, and so this is not a vector space.
Note, the mathematical intuition to recognize if a set with two operations is probably not a vector space—and the skill in finding a suitable counterexample—are both developed through practice.
Example: Let S be the set of all 2 by 2 matrices [a1, a2; a3, a4] such that a1 + a2 + a3 + a4 = 0, and a1, a2, a3, and a4 are real numbers. Is S a vector space under standard addition and scalar multiplication of matrices? Unlike the last example, we see that we are using standard operations, and so these should work. However, the set itself is a little different, so we must be careful about the interaction between the operations and the set. Recall way back to Theorem 1.1.1. When we first discussed the ten properties of vectors in Rn, we observed that properties V1, V4, V5, and V6 were about the relationship between the operations and the set, while the other properties were only about the operations. Indeed, since we are using standard addition and scalar multiplication of matrices, Theorem 3.1.1 tells us that properties V2, V3, V7, V8, V9, and V10 must hold. Therefore, we only really have to check the remaining four properties. Let A = [a1, a2; a3, a4] and B = [b1, b2; b3, b4] be matrices in S. Then by definition of S, we have a1 + a2 + a3 + a4 = 0, and b1 + b2 + b3 + b4 = 0. Property V1: Observe that A + B = [a1 + b1, a2 + b2; a3 + b3, a4 + b4]. If we sum all of the components of A + B, we get (a1 + b1) + (a2 + b2) + (a3 + b3) + (a4 + b4), which we can rearrange to be (a1 + a2 + a3 + a4) + (b1 + b2 + b3 + b4), which is just 0 + 0, which is 0. Therefore, A + B satisfies the condition of S, and hence, A + B is in S. For property V4, we need to pick the zero vector for the set. Since we are using the same operations as the vector space M(2 by 2)(R), of 2 by 2 matrices with real entries, the zero vector must be the same. That is, the zero vector would have to be the 2 by 2 zero matrix. For V4 to be fully satisfied, the zero vector must belong to the set S. But the sum of the components of the zero matrix is 0, and so S has a zero vector. V5: Using the same thinking process as for V4, the additive inverse of A = [a1, a2; a3, a4] has to be –A = [-a1, -a2; -a3, -a4]. –A is in S since (-a1) + (-a2) + (-a3) + (-a4) = (-1)(a1 + a2 + a3 + a4), which is (-1)(0) = 0. Finally, for V6, we have that for any real scalar t, tA = [ta1, ta2; ta3, ta4], which is in S since ta1 + ta2 + ta3 + ta4 = t(a1 + a2 + a3 + a4) = t(0) = 0. Therefore, all ten axioms hold, and so S is a vector space under standard addition and scalar multiplication of matrices.
This should remind you a lot of what we did for subspaces of Rn. Moreover, by definition of a subspace of Rn, we see that every subspace of Rn is, in fact, a vector space. We generalize the concept of subspaces to general vector spaces.
Definition: If S is a non-empty subset of a vector space V, and S is also a vector space using the same operations as V, then S is called a subspace of V. Note that in the previous example, we saw that we didn’t have to check some of the axioms because we were using operations from something we already knew was a vector space. So just as we saw for subspaces of Rn, when checking if a non-empty subset S is a subspace of a vector space V, we only need to check properties V1 and V6 since we are using the same operations as V. We again state this as the Subspace Test.
The Subspace Test: A non-empty subset S of a vector space V is a subspace of V if S is closed under addition and closed under scalar multiplication under the operations of V. Notes: First, the proof of the Subspace Test is essentially the same as the proof of the Subspace Test in Rn. Second, it is important not to forget to show that S is a non-empty subset of V. As with subspaces of Rn, the best way to show the set is non-empty is to show that the zero vector of V is in S.
Example: Let V be a vector space. Show that V, and the set U which only contains the zero vector of V, is a subspace of V. First, notice that V is a subset of itself, and is a vector space under its own operations, and so it satisfies the definition of a subspace. Therefore, every vector space is a subspace of itself. For U, by definition, it is a non-empty subset of V, so we pick any two vectors x and y in U. Hmm, the only vector in U is 0, and so x = 0 and y = 0. For V1, we have x + y = 0 + 0. The important thing to notice here (and for the rest of this example) is that we are using the same operations as V, and we already know that V is a vector space. Hence, we know all ten properties hold for V. So, we can apply property V4 of V to say that 0 + 0 = 0. Therefore, x + y is in U. Proving V6 is a little trickier. We need to show that any scalar multiple of x is in U. In other words, we need to prove that any scalar multiple of 0 is 0. We have t0 = t(0x) by Theorem 4.1.1. Using property V7 of V, this equals (t(0))x. t(0) is just 0, and applying Theorem 4.1.1 again, we get that 0x = 0. Hence, by the Subspace Test, U is a subspace of V. Note that by definition of a subspace, this means the set U, which only contains 0V, is itself a vector space under the operations of V.
In general, if we were trying to prove a set S is a vector space, and we already know that S is a subset of a vector space V and uses the same operations as V, then we just need to prove that S is a subspace of V to prove that S is a vector space.
Example: Let W equal the set of all polynomials in P3(R) such that 1 is a root of p. Show that W is a vector space under standard addition and scalar multiplication of polynomials. As just mentioned, we can prove that W is a vector space by showing that W is a subspace of P3(R). So, by definition, we see that W is a subset of P3(R). Also, the zero vector in P3(R) is the zero polynomial z(x) = 0, which satisfies z(1) = 0, and so the zero polynomial is in W, and hence W is non-empty. Let p and q be polynomials in W. Then by definition of W, we have p(1) = 0 and q(1) = 0. For V1, we have (p + q)(1), by definition of addition of polynomials, is p(1) + q(1), which is 0 + 0, which equals 0. Therefore, p + q is in W. Similarly, for V6, we have that (tp)(1) = tp(1) = t(0) = 0. Therefore, tp is also in W. Consequently, W is a subspace of P3(R) by the Subspace Test, and hence, W is itself a vector space.
Example: Let S be the set of polynomials ax-squared + bx + c in P2(R) such that a-squared – b-squared = 0. Is S a subspace of P2(R)? Before we immediately begin trying to apply the Subspace Test, we notice that the condition on the set involves squares. This makes us think that the set is probably not a subspace. In particular, squaring the variables should mean that the set is not closed under addition. To prove this, of course, we need to find a counterexample. Observe that x-squared + x and 2x-squared – 2x are in S, since 1-squared – 1-squared = 0 and 2-squared – (-2)-squared = 0, but their sum (x-squared + x) + (2x-squared – 2x) = 3x-squared – x is not in S, since 3-squared – (-1)-squared is not equal 0. So S is not closed under addition, and hence, S is not a subspace.
Example: Let S be the set of 2 by 2 matrices in M(2 by 2)(R) such that x1 + x2 = 2x4. Is S a subspace of M(2 by 2)(R)? We first observe that, by definition, S is a subset of M(2 by 2)(R), and the zero matrix satisfies the condition of S since 0 + 0 = 2(0). Therefore, the zero matrix is in S, and hence, S is non-empty. Let A and B be any two matrices in S. Then A has the form [a1, a2; a3, a4] where a1 + a2 = 2a4, and B has the form [b1, b2; b3, b4] where b1 + b2 = 2b4. For V1, we get A + B = [a1 + b1, a2 + b2; a3 + b3, a4 + b4]. Is this in S? We have to check if A + B satisfies the conditions of the set S. We observe that (a1 + b1) + (a2 + b2) = (a1 + a2) + (b1 + b2) = 2a4 + 2b4 = 2(a4 + b4), so A + B does satisfy the condition of S, and so A + B is in S. Similarly, for V6, we have tA = [ta1, ta2; ta3, ta4] where ta1 + ta2 = t(a1 + a2) = t(2a4), which can be rewritten as 2(ta4), and so tA is in S. Thus, S is a subspace of M(2 by 2)(R) by the Subspace Test.
As you can see, the Subspace Test is very algorithmic. With enough practice, you should become quick at identifying if a set is or isn’t a subspace, and fast at either proving the set is a subspace with the Subspace Test or at finding a counterexample to prove the set isn’t a subspace.
This concludes this lecture. In the next lecture, we will continue to mimic what we did with vectors in Rn. In particular, we will look at the very important concepts of spanning and linear independence.