# Chapter 4 Vector Spaces - University of Kansas

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Chapter 4
Vector Spaces
4.1 Vectors in Rn
Homework: [Textbook, §4.1 Ex. 15, 21, 23, 27, 31, 33(d), 45, 47, 49, 55, 57; p. 189-]. We discuss vectors in plane, in this section.
In physics and engineering, a vector is represented as a directed segment. It is determined by a length and a direction. We give a short review of vectors in the plane.
Deﬁnition 4.1.1 A vector x in the plane is represented geometrically by a directed line segment whose initial point is the origin and whose terminal point is a point (x1, x2) as shown in in the textbook,
115

116 page 180.

CHAPTER 4. VECTOR SPACES

y
O

~? •

x

~ ~

~

~

~

~

/x

The bullet at the end of the arrow is the terminal point (x1, x2). (See the textbook,page 180 for a better diagram.) This vector is represented by the same ordered pair and we write
x = (x1, x2).
1. We do this because other information is superﬂuous. Two vectors u = (u1, u2) and v = (v1, v2) are equal if u1 = v1 and u2 = v2.
2. Given two vectors u = (u1, u2) and v = (v1, v2), we deﬁne vector addition u + v = (u1 + v1, u2 + v2). See the diagram in the textbook, page 180 for geometric interpretation of vector addition.
3. For a scalar c and a vector v = (v1, v2) deﬁne
cv = (cv1, cv2)
See the diagram in the textbook, page 181 for geometric interpretation of scalar multiplication.
4. Denote −v = (−1)v.

4.1. VECTORS IN RN

117

Reading assignment: Read [Textbook, Example 1-3, p. 180-] and study all the diagrams.
Obvioulsly, these vectors behave like row matrices. Following list of properties of vectors play a fundamental role in linear algebra. In fact, in the next section these properties will be abstracted to deﬁne vector spaces.

Theorem 4.1.2 Let u, v, w be three vectors in the plane and let c, d

be two scalar.

1. u + v is a vector in the plane closure under addition

2. u + v = v + u

3. (u + v) + w = u + (v + w) Associate property of addition

4. (u + 0) = u

5. u + (−1)u = 0

6. cu is a vector in the plane closure under scalar multiplication

7. c(u + v) = cu + cv

Distributive propertyof scalar mult.

8. (c + d)u = cu + du

Distributive property of scalar mult.

9. c(du) = (cd)u

Associate property of scalar mult.

10. 1(u) = u

Multiplicative identity property

Proof. Easy, see the textbook, papge 182.

4.1.1 Vectors in Rn
The discussion of vectors in plane can now be extended to a discussion of vectors in n−space. A vector in n−space is represented by an ordered n−tuple (x1, x2, . . . , xn).
The set of all ordered n−tuples is called the n−space and is denoted by Rn. So,
1. R1 = 1 − space = set of all real numbers,

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2. R2 = 2 − space = set of all ordered pairs (x1, x2) of real numbers
3. R3 = 3 − space = set of all ordered triples (x1, x2, x3) of real numbers
4. R4 = 4 − space = set of all ordered quadruples (x1, x2, x3, x4) of real numbers. (Think of space-time.)
5. . . . . . .
6. Rn = n−space = set of all ordered ordered n−tuples (x1, x2, . . . , xn) of real numbers.
Remark. We do not distinguish between points in the n−space Rn and vectors in n−space (deﬁned similalry as in deﬁnition 4.1.1). This is because both are describled by same data or information. A vector in the n−space Rn is denoted by (and determined) by an n−tuples (x1, x2, . . . , xn) of real numbers and same for a point in n−space Rn. The ith−entry xi is called the ith−coordinate.
Also, a point in n−space Rn can be thought of as row matrix. (Some how, the textbook avoids saying this.) So, the addition and scalar multiplications can be deﬁned is a similar way, as follows.

Deﬁnition 4.1.3 Let u = (u1, u2, . . . , un) and v = (v1, v2, . . . , vn) be vectors in Rn. The the sum of these two vectors is deﬁned as the vector
u + v = (u1 + v1, u2 + v2, . . . , un + vn). For a scalar c, deﬁne scalar multiplications, as the vector
cu = (cu1, cu2, . . . , cun). Also, we deﬁne negative of u as the vector
−u = (−1)(u1, u2, . . . , un) = (−u1, −u2, . . . , −un) and the diﬀerence
u − v = u + (−v) = (u1 − v1, u2 − v2, . . . , un − vn).

4.1. VECTORS IN RN

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Theorem 4.1.4 All the properties of theorem 4.1.2 hold, for any three vectors u, v, w in n−space Rn and salars c, d.

Theorem 4.1.5 Let v be a vector in Rn and let c be a scalar. Then,

1. v + 0 = v. (Because of this property, 0 is called the additive identity in Rn.) Further, the additive identitiy unique. That means, if v + u = v for all vectors v in Rn than u = 0.
2. Also v + (−v) = 0. (Because of this property, −v is called the additive inverse of v.) Further, the additive inverse of v is unique. This means that v + u = 0 for some vector u in Rn, then u = −v.
3. 0v = 0. Here the 0 on left side is the scalar zero and the bold 0 is the vector zero in Rn.
4. c0 = 0.
5. If cv = 0, then c = 0 or v = 0.
6. −(−v) = v.
Proof. To prove that additive identity is unique, suppose v + u = v for all v in Rn. Then, taking v = 0, we have 0 + u = 0. Therefore, u = 0.
To prove that additive inverse is unique, suppose v+u = 0 for some vector u. Add −v on both sides, from left side. So,
−v + (v + u) = −v + 0

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So, (−v + v) + u = −v
So, 0 + u = −v So, u = −v.

We will also prove (5). So suppose cv = 0. If c = 0, then there is nothing to prove. So, we assume that c = 0. Multiply the equation by c−1, we have c−1(cv) = c−10. Therefore, by associativity, we have (c−1c)v = 0. Therefore 1v = 0 and so v = 0.
The other statements are easy to see. The proof is complete.
Remark. We denote a vector u in Rn by a row u = (u1, u2, . . . , un). As I said before, it can be thought of a row matrix

u = u1 u2 . . . un .

In some other situation, it may even be convenient to denote it by a

column matrix:

 u1 

u

=

 

u2

 .

 ... 

un

Obviosly, we cannot mix the two (in fact, three) diﬀerent ways.

Exercise 4.1.6 (Ex. 46, p. 189) Let u = (0, 0, −8, 1) and v = (1, −8, 0, 7). Find w such that 2u + v − 3w = 0.

Solution: We have

w = 2u + 1v = 2(0, 0, −8, 1) + 1(1, −8, 0, 7) = (1, −8 , −16, 3).

33 3

3

33 3

Exercise 4.1.7 (Ex. 50, p. 189) Let u1 = (1, 3, 2, 1), u2 = (2, −2, −5, 4), u3 = (2, −1, 3, 6). If v = (2, 5, −4, 0), write v as a linear combination of u1, u2, u3. If it is not possible say so.

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Solution: Let v = au1 +bu2 +cu3. We need to solve for a, b, c. Writing the equation explicitly, we have

(2, 5, −4, 0) = a(1, 3, 2, 1) + b(2, −2, −5, 4) + c(2, −1, 3, 6).

Therefore

(2, 5, −4, 0) = (a + 2b + 2c, 3a − 2b − c, 2a − 5b + 3c, a + 4b + 6c)

Equating entry-wise, we have system of linear equation

a +2b +2c = 2 3a −2b −c = 5 2a −5b +3c = −4 a +4b +6c = 0

We write the augmented matrix:

1 2 22

 

3

−2

−1

5

 

 2 −5 3 −4 

1 4 60

We use TI, to reduce this matrix to Gauss-Jordan form:

1002

 

0

1

0

1

 

 0 0 1 −1 

0000

So, the system is consistent and a = 2, b = 1, c = −1. Therefore

v = 2u1 + u2 − u3,

which can be checked directly,

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4.2 Vector spaces

Homework: [Textbook, §4.2 Ex.3, 9, 15, 19, 21, 23, 25, 27, 35; p.197].

The main pointin the section is to deﬁne vector spaces and talk about examples.

The following deﬁnition is an abstruction of theorems 4.1.2 and theorem 4.1.4.
Deﬁnition 4.2.1 Let V be a set on which two operations (vector addition and scalar multiplication) are deﬁned. If the listed axioms are satisﬁed for every u, v, w in V and scalars c and d, then V is called a vector space (over the reals R).
(a) u + v is a vector in V (closure under addition). (b) u + v = v + u (Commutative property of addition ). (c) (u + v) + w = u + (v + w) (Associative property of addition). (d) There is a zero vector 0 in V such that for every u in V
we have (u + 0) = u (Additive identity). (e) For every u in V , there is a vector in V denoted by −u such
that u + (−u) = 0 (Additive inverse).
2. Scalar multiplication:
(a) cu is in V (closure under scalar multiplication0.

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123

(b) c(u + v) = cu + cv (Distributive propertyof scalar mult.). (c) (c + d)u = cu + du (Distributive property of scalar mult.). (d) c(du) = (cd)u (Associate property of scalar mult.). (e) 1(u) = u (Scalar identity property).

Remark. It is important to realize that a vector space consisits of four entities:
1. A set V of vectors.
2. A set of scalars. In this class, it will alawys be the set of real numbers R. (Later on, this could be the set of complex numbers C.)
3. A vector addition denoted by +.
4. A scalar multiplication.

Lemma 4.2.2 We use the notations as in deﬁnition 4.2.1. First, the zero vector 0 is unique, satisfying the property (1d) of deﬁnition 4.2.1.
Further, for any u in V , the additive inverse −u is unique.
Proof. Suppose, there is another element θ that satisfy the property (1d). Since 0 satisfy (1d), we have
θ = θ + 0 = 0 + θ = 0.
The last equality follows because θ satisfes the property(1d). (The proof that additive inverse of u unique is similar the proof
of theorem 2.3.2, regarding matrices.) Suppose v is another additive inverse of u.
u + v = 0 and u + (−u) = 0.

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So.
−u = 0 + (−u) = (u + v) + (−u) = v + (u + (−u)) = v + 0 = v.
So, the proof is complete.
Reading assignment: Read [Textbook, Example 1-5, p. 192-]. These examples lead to the following list of important examples of vector spaces:

Example 4.2.3 Here is a collection examples of vector spaces:
1. The set R of real numbers R is a vector space over R.
2. The set R2 of all ordered pairs of real numers is a vector space over R.
3. The set Rn of all ordered n−tuples of real numersis a vector space over R.
4. The set C(R) of all continuous functions deﬁned on the real number line, is a vector space over R.
5. The set C([a, b])) of all continuous functions deﬁned on interval [a, b] is a vector space over R.
6. The set P of all polynomials, with real coeﬃcients is a vector space over R.
7. The set Pn of all polynomials of degree ≤ n, with real coeﬃcients is a vector space over R.
8. The set Mm,n of all m × n matrices, with real entries, is a vector space over R.