Lesson: Vector Spaces

Question 2

1 point

Transcript — Introduction

In this lecture, we will define the abstract concept of a real vector space, and show a variety of examples of vector spaces.

Definition: A real vector space is a set V, together with an operation of addition, denoted x + y, and an operation of scalar multiplication, denoted tx for any real number t, such that for any x, y, and z in V, and real numbers a and b, we have all of the following properties:

  1. V1: x + y is in V
  2. V2: (x + y) + z = x + (y + z)
  3. V3: x + y = y + x
  4. V4: There exists a vector 0 in V, called the zero vector, such that x + 0 = x.
  5. V5: There exists a vector –x in V such that x + (-x) = 0.
  6. V6: ax is in V
  7. V7: a(bx) = (ab)x
  8. V8: (a + b)x = ax + bx
  9. V9: a(x + y) = ax + ay
  10. V10: 1x = x

At this point, you should have all ten of these properties memorized, as you have seen them quite a few times before. If you don’t have them memorized, it is important that you memorize them now, as you will need them to do certain problems involving vector spaces.

We can actually generalize the concept of a vector space even further by allowing the scalars to be from any field, such as the rational numbers or the complex numbers. However, for purposes of this course, we will just look at real vector spaces—that is, vector spaces where the scalars are only allowed to be real numbers. So, in this course, whenever we say “vector space”, we mean “real vector space”.

Note, we sometimes denote addition in a vector space by plus with a circle around it, and scalar multiplication by a dot with a circle around it, to stress that the operations are non-standard. This will be demonstrated in a couple of examples later.

In some cases, we will denote the zero vector of a vector space V by 0-subscript-V to stress which vector space it belongs to.

We now look at several examples of vector spaces. Example: Rn is a vector space with addition and scalar multiplication defined in the usual way. We call these standard addition and standard scalar multiplication. Whenever we are looking at a vector space, it is important that we identify the zero vector and the additive inverses of any vector x. In Rn, we have seen that 0 is the vector with every component 0, and the additive inverse of x is –x = (-1)x.

Example: M(m by n)(R), the set of all m by n matrices, is a vector space with standard addition and scalar multiplication of matrices. We have seen that 0 is the zero matrix, and the additive inverse of the matrix A is –A = (-1)A.

Example: The set L of all linear mappings L from Rn to Rm with standard addition and scalar multiplication of linear mappings is a vector space. The zero vector is the linear mapping 0 from Rn to Rm such that 0(x) = 0 in Rm for all x. The additive inverse of L is –L = (-1)L.

Example: The set Pn(R) of all polynomials of degree at most n with real coefficients is a vector space with standard addition and scalar multiplication of polynomials. The zero vector in Pn(R) is the zero polynomial, 0 = 0 + 0x + up to 0(x-to-the-n). The additive inverse of the polynomial p(x) = a0 + a1x + up to an(x-to-the-n) is (-p)(x) = (-1)p(x) = -a0 – a1x - up to an(x-to-the-n).

At this point, we pause to stress something very important about the definition of a vector space. Notice in the definition that it specifies a vector space is a set V with two operations of addition and scalar multiplication that satisfy the ten properties. Notice, in the previous four examples, we have been very careful to not only identify the set of vectors that we are using, but also to identify what operations of addition and scalar multiplication are being used in those vector spaces. As we will see, it is possible for us to change the definitions of addition and scalar multiplication even if we are using the same set. This set with these two new operations of addition and scalar multiplication will be a vector space only if they satisfy all ten properties. We will look at some examples like this in the next lecture.

For now, as previously mentioned, we are defining vector spaces to have the same structure as Rn. The study of vector spaces is the study of this common structure. However, it is possible that vectors in individual vector spaces have other aspects not common to all vector spaces, such as matrix multiplication in M(m by n)(R) and factorization of polynomials in Pn(R).

So now let’s look at an example of this common structure. Notice in all of the examples of vector spaces we just gave that we have the scalar 0 times any vector is the zero vector in the vector space, and that -1 times any vector v is the additive inverse of v. Since all vector spaces have the same structure, these properties should hold for all vector spaces. Theorem 4.1.1: If V is a vector space, and v is any vector in V, then

  1. The zero-vector in V = (0)v
  2. The additive inverse of v = (-1)v

We prove (1) and leave (2) as an exercise. In the examples we have looked at so far, these properties were fairly obvious. However, to prove it is true for all vector spaces, no matter how strange the set and operations of addition and scalar multiplication, we can only use the ten properties from the definition of a vector space. We can take nothing for granted. The difficulty with this type of proof at first is making sure that you don’t skip steps. Okay, so here we go. Proof: For all vectors v in V, we have 0v = 0v + 0 by using property V4. By property V5, we have 0 = v + its additive inverse, so we have 0v + (v + the additive inverse of v). By property V10, v = 1v, so we get 0v + (1v + the additive inverse of v). We then use property V2, which says this is equal to (0v + 1v) + the additive inverse of v. We use property V8 to get that this is equal to (0 + 1)v + the additive inverse of v. 0 and 1 are just real numbers, so we can add them like normal, so we get 1v + the additive inverse of v. Yes, we have to use property V10 again, to say that 1v is indeed v, so we have this is equal to v + the additive inverse of v, which, by property V5, is 0. And so 0v does in fact equal 0 for every v in V.

This theorem proves a few things. It shows that we can find the zero vector in any vector space by picking any vector v in the vector space and multiplying it by the scalar 0. Notice that this implies that the zero vector in a vector space is unique. The theorem also demonstrates that the additive inverse of any vector v in the vector space is unique, and equals to (-1)v. We will see that this theorem can help us when working with more complicated vector spaces.

And now let’s do an example of a strange vector space. Let D be the set of all positive real numbers. We define non-standard addition in the set D by x + y = xy, and we define scalar multiplication in D by tx = x-to-the-power-of-t. We will prove that D, with these two operations, is in fact a vector space. Pick any vectors x, y, and z in D, and real scalars c and d. By definition of D, since x, y, and z are in D, we have that x, y, and z are all positive real numbers.

We now need to show that all ten properties hold. To prove property V1, we need to show that D is closed under addition. By definition, we have that x + y = xy. We need to show that this element belongs to the set D. Since x and y are both positive, we get xy is positive, and hence satisfies the conditions on the set D. Thus, x + y is in D.

For V2, we need to show addition is associative. We start with (x + y) + z. By order of operations, we do what’s inside the brackets first. By definition of plus, this gives (xy) + z. Notice that this is well-defined, since we just proved that xy is a vector in D. Now by definition of plus, (xy) + z = (xy)z. Now this is just normal multiplication of real numbers, which is associative, so we can write this as x(yz). By definition of plus in D, we get x + (yz), which is x + (y + z) as required.

For V3, we show that addition is commutative. We have x + y = xy = yx, which, by definition, is y + x.

For V4, we need to show the existence of a zero vector in the set. But what is the zero vector? This is where Theorem 4.1.1 helps out. Theorem 4.1.1 tells us that if D is a vector space, then the zero vector must be 0x. In the case of D, we get 0x is x-to-the-power-of-0, which is 1 since x is not equal to 0. Thus, if D is a vector space, then the zero vector must be 1, so we now prove that 1 is indeed the zero vector for D. First, we observe that 1 does belong to the set D since it is a positive real number. Also, it satisfies x + 1 = x(1), which is x for all x in D. Therefore, the zero vector is 1. Note, this is one of the few times that you can write 0 = 1 and be correct.

For V5, we need to prove the existence of additive inverses. Again, we make use of Theorem 4.1.1. It says that if D is a vector space, then the additive inverse of any x in D is (-1)x, which in this case is x-to-the-power-of-(-1), which is 1/x. We now prove that 1/x is indeed the additive inverse for x. First, 1/x is greater than 0 since x is greater than 0, and hence 1/x is in the set D. Also, 1/x + x = x(1/x) = 1, which is the zero vector. Thus, 1/x is the additive inverse for x.

For V6, we need to prove that D is closed under scalar multiplication. We have that cx = x-to-the-power-of-c, which is greater than zero since x is greater than 0. Thus, xc is in D.

For V7, we prove that scalar multiplication is associative. We get that c(dx), by definition of scalar multiplication, equals c(x-to-the-power-of-d), which, again using the definition of scalar multiplication, is (x-to-the-power-of-d)-all-to-the-power-of-c. Using properties of exponents, this is x-to-the-power-of-(cd), which, by definition of scalar multiplication, is (cd)x.

For V8, we prove one of the distributive laws. We have that (c + d)x = x-to-the-power-of-(c + d). By properties of exponents, this is (x-to-the-power-of-c)(x-to-the-power-of-d). We now use the definition of addition in D to get this is equal to x-to-the-power-of-c + x-to-the-power-of-d. Finally, using the definition of scalar multiplication, we get this equals (cx) + (dx).

For V9, we prove another distributive law. We have c(x + y) , by definition of addition, is c(xy), which by definition of scalar multiplication is (xy)-to-the-power-of-c. This is (x-to-the-power-of-c)(y-to-the-power-of-c), which by definition of addition is x-to-the-power-of-c + y-to-the-power-of-c, and finally this gives (cx) + (cy).

Last, and probably least, V10: 1 is the scalar multiplicative identity. We have 1x = x-to-the-power-of-1, which equals x.

Since we have proven all ten properties hold, D is a vector space with this definition of addition and scalar multiplication. Take a few minutes to look over and make sure you understand the proof and all of the steps. Take careful note to notice the difference between the scalars (c and d) and the vectors (x, y, and z) and how we’re using the different definitions for addition and scalar multiplication.

At first glance, it seems strange that this set D, with these weird definitions for addition and scalar multiplication, has the same structure as the real numbers. However, observe that we can relate numbers in R to numbers in D by any number x in R is the vector 2-to-the-power-of-x in D. That is, the number 0 in R corresponds to the vector 2-to-the-power-of-0 which equals 1 in D, and the addition 1 + 2 = 3 corresponds to the addition in D, 2 + 4 = 8. This is because the real number 1 corresponds to 2 = 2-to-the-power-of-1, the real number 2 corresponds to 2-to-the-power-of-2 = 4, and the real number 3 corresponds to 2-to-the-power-of-3 = 8.

This concludes this lecture. In the next lecture, we will look at some more examples of vector spaces, and then define a subspace of a vector space, just as we did for Rn.

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