Chapter 2
Vector Spaces
One of my favorite dictionaries (the one from Oxford) defines a vector as “A quantity having direction as well as magnitude, denoted by a line drawn from its original to its final position.” What is useful about this definition is that we can draw pictures and use our spatial intuition. An objection to this being used as the definition of a vector is: It’s not very precise, and thus will be hard to compute with to any degree of accuracy.
The first section of this chapter makes the phrase “a quantity having direc- tion and magnitude” more precise and at the same time develops an algebraic structure. That is, the addition of one vector to another and the multiplication of vectors by numbers will be defined, and various properties of these operations will be stated and proved.
In order to avoid confusing vectors with numbers, all vectors will be in boldface type.
2.1 R
2through R
nFix some point (think of it as the origin in the Euclidean plane), draw two short rays from this point, and put arrowheads at the tips of the rays (see Figure 2.1).
B A
Figure 2.1
You’ve just drawn two vectors. Now label one of them AAA and the other BBB. The magnitudes of these vectors are their lengths.
In many applications, vectors are used to represent forces. Since several forces may act on an object, and the result of this will be the same as if a single
55
force, called the resultant or sum, acted on the object, we define the sum of two vectors in order to model how forces combine. Thus, if we want the vector AAA + BBB, we draw a dashed line segment starting at the tip of AAA, parallel to BBB, with the same length as BBB; cf. Figure 2.2a and b. Label the end of this dashed line c and now draw the vector CCC; i.e., draw a line segment from the origin to c and put an arrowhead there; cf. Figure 2.2a. The vector CCC is called AAA + BBB, and this way of combining AAA and BBB is referred to as the parallelogram law of vector addition. Note that the construction of BBB + AAA will be different from that of AAA + BBB; cf. Figure 2.2b. However, a little geometry convinces us that we get the same two vectors. We repeat this. The fact that AAA + BBB = BBB + AAA is something that has to be proved. Its truth is not self-evident.
There is another operation (scalar multiplication) that we can perform on a vector, and that is to multiply it by a number. If AAA is a vector, 2AAA’s meaning is clear. It is the same as AAA + AAA; i.e., 2AAA is a vector twice as long, and with the same direction, as AAA. Thus, we define cAAA (c ≥ 0) to be the vector pointing in the same direction as AAA but whose magnitude is c times the magnitude of AAA. If c is negative, cAAA points in the direction opposite to AAA.
Since we will be doing a lot of computing with vectors, we need a method for doing so other than using a ruler, compass, and protractor. Following in the footsteps of Descartes and others, we assign a pair of numbers to each vector and then see how these pairs should be combined in order to model vector addition and scalar multiplication.
A B + A
A + B
B A
(b) (a)
B
Figure 2.2
Let’s go back to Figure 2.1. This time we label our initial point (0,0), the origin in the Euclidean plane. We also draw two perpendicular lines that intersect at (0,0). We call the horizontal line the x1 axis and the vertical line the x2 axis. We next associate a number with each point on these axes. This number indicates the directed distance of the point from the origin; that is, the point on the x1 axis labeled 2 is 2 units to the right of the origin while the point labeled −2 is 2 units to the left. How large a distance the number 1 represents is arbitrary. For the x2 axis, a positive number indicates that the point lies above the origin while a negative number means that the point lies below the origin. We now associate an ordered pair of numbers with each point in the plane. The first number tells us the directed distance of the point from the x2 axis along a line parallel to the x1 axis and the second number gives us the directed distance of the point from the x1 axis along a line parallel to the
2.1. R2 THROUGH RN 57 x2 axis; cf. Figure 2.5. Every vector, which we picture as an arrow emanating from the origin, is uniquely determined once we know where its tip is located.
This means that every vector can be uniquely associated with an ordered pair of numbers. In other words, two vectors are equal if and only if their respective coordinates are the same.
(a) (b) (c)
A
A A
A c , c> 0
c c> 0
2 A
A,
Figure 2.3
The preceding discussion may lead to some confusion as to what an ordered pair of numbers represents: a point in the plane or a vector. In practice this will not cause any problem, since one can just as easily think of a vector as a point in the plane, that point where the tip of the vector is located.
Example 1. Sketch the following vectors:
a. (1,1)
b. (−1, 3)
c. (2, −3)
Our next task is to determine how the number pairs, i.e., the coordinates, of two vectors should be combined to give their sum. Let AAA = (a1, a2) and BBB = (b1, b2) be two vectors. Then a simple proof using congruent triangles yields that AAA + BBB = (a1+ b1, a2+ b2); cf. Figure 2.6a. Similar arguments show us that cAAA = (ca1, ca2) if c is a rational number; cf. Figure 2.6b. We haven’t talked about subtracting one vector from another yet, but AAA−BBB should be equal to a vector such that (AAA − BBB) + BBB = AAA. Thus if AA = (aA 1, a2) and BBB = (b1, b2), then AAA − BBB = (a1− b1, a2− b2).
x1
(0,0) (2,0) (−2, 0)
(0,1) x2
Figure 2.4
b < 0 a < 0
(a, 0) (a, b) b > 0
a < 0
x2
(0, b)
a > 0 b > 0
a > 0 b < 0
x1
Figure 2.5
We now formally define R2along with vector addition and scalar multiplica- tion for these particular vectors. R2is the classical example of a two-dimensional vector space.
2.1. R2 THROUGH RN 59
(a1, a2)
(b) (a)
(a1+ b1, a2+ b2) (a1, a2)
(ca1, ca2) a2
a1+ b1
a1 b1
(b1, b2) b2
a2+ b2
Figure 2.6
Definition 2.1. R2 = {(x1, x2) : x1 and x2 are arbitrary real numbers). If AA
A = (a1, a2), BBB = (b1, b2), and c is any real number, then 1. AAA + BBB = (a1+ b1, a2+ b2)
2. cAAA = (ca1, ca2)
Note that R2 can also be thought of as all possible 1 × 2 matrices, with vector addition being matrix addition and scalar multiplication the same as multiplying a matrix by a scalar; cf. Definitions 1.3 and 1.4.
A few words about the notation {: }. It is used to describe a set of elements.
The phrase after the colon describes the property or attributes that something must possess in order for it to be in the set. For example, {x: x = 1, 2} is the set {1, 2}, and {x: 1 ≤ x < 100} is the set of real numbers between 1 and 100 including 1 but excluding 100. Thus (1,23), (3.14159,53), and (0, −7) are in R2, but (0,$), (#, f ), and !e, are not, since the last three objects are not pairs of real numbers.
Example 2. Solve the following vector equation:
2(6, −1) + 3XXX = (3, −4)
3XXX = (3, −4) − 2(6, −1)
= (3, −4) − (12, −2)
= (3 − 12, −4 + 2) = (−9, −2)
Thus
XXX = 1
3(−9, −2) =
−3, −2 3
Example 3. Given any vector in R2write it as the sum of two vectors that are parallel to the coordinate axes.
Solution. Let AAA = (a1, a2) be any vector in R2. To say that a vector XXX is parallel to the x1 axis is to say that the second component of XXX’s representation as an ordered pair of numbers is zero. A similar comment applies to vectors parallel to the x2 axis. Thus,
AAA = (a1, a2) = (a1, 0) + (0, a2)
The first vector, (a1, 0), is parallel to the x1 axis and the second vector, (0, a2), is parallel to the x2 axis. We could also write
A
AA = (a1, 0) + (0, a2) = a1(1, 0) + a2(0, 1)
The vectors (1,0) and (0,1) are often denoted by iii and jjj, respectively, and in
this notation (a1, a2) = a1iii + a2jjj.
(a1, a2)
a1iii= (a1,0) a2jjj= (0, a2)
Figure 2.7
The theorem below lists some of the algebraic properties that vector addition and scalar multiplication in R2satisfy.
Theorem 2.1. Let AAA = (a1, a2), BBB = (b1, b2), and CCC = (c1, c2) be three arbitrary vectors in R2. Let a and b be any two numbers. Then the following equations are true.
1. AAA + BBB = BBB + AAA
2. (AAA + BBB) + CCC = AAA + (BBB + CCC)
3. Let 000 = (0, 0), then AAA + 000 = AAA [zero vector]
4. For every AAA there is a (−AAA) such that AAA + (−AAA) = 000[−AAA = (−a1, −a2)]
5. a(AAA + BBB) = aAAA + aBBB
2.1. R2 THROUGH RN 61 6. (a + b)AAA = aAAA + bAAA
7. (ab)AAA = a(bAAA) 8. 1AAA = AAA
Proof. We verify equations 1, 4, 5, and 8, leaving the others for the reader.
1. AAA + BBB = (a1, a2) + (b1, b2) = (a1+ b1, a2+ b2) = (b1+ a1, b2+ a2)
= (b1, b2) + (a1, a2) = BBB + AAA
4. AAA + (−AAA) = (a1, a2) + (−a1, −a2) = (a1− a1, a2− a2) = (0, 0) = 000 5. a(AAA + BBB) = a[(a1, a2) + (b1, b2)] = a(a1+ b1, a2+ b2)
= (a(a1+ b1), a(a2+ b2)) = (aa1+ ab1, aa2+ ab2)
= (aa1, aa2) + (ab1, ab2) = a(a1, a2) + a(b1, b2)
= aAAA + aBBB
8. 1AAA = 1(a1, a2) = (1a1, 1a2) = (a1, a2) = AAA
We next discuss the standard three-dimensional space R3. As with R2, we first picture arrows starting at some fixed point and ending at any other point in three-dimensional space. We draw three mutually perpendicular coordinate axes x1, x2, and x3 and impose a distance scale on each of the axes. To each point P in three space we associate an ordered triple of numbers (a, b, c), where a denotes the directed distance from P to the x2, x3plane, b the directed distance from P to the x1, x3 plane, and c the directed distance from P to the x1, x2
plane. Just as we did for R2, we now think of vectors in three space both as arrows and as ordered triples of numbers.
Example 4. For each of the triples of numbers below sketch the vector they represent.
a. (1,0,0), (0,1,0), (0,0,1): these three vectors are commonly denoted by iii, jjj, and kkk, respectively.
jjj= (0, 1, 0)
x2
iii= (1, 0, 0) x3
x1
k k
k= (0, 0, 1)
This notation is somewhat ambiguous; e.g., does iii represent (1,0) or (1,0,0)? There should, however, be no confusion, since which vector space we are talking about will be clear from the discussion, and this will deter- mine the meaning of iii.
b. AAA = (−1, 2, 1)
(−1, 2, 0) (−1, 0, 0) (−1, 2, 1)
x1 x3
x2 (0, 2, 0)
Definition 2.2 defines R3 and the algebraic operations of vector addition and scalar multiplication for this vector space.
Definition 2.2. R3 = {(x1, x2, x3) : x1, x2, and x3 are any real numbers}.
Vector addition and scalar multiplication are defined as follows:
1. AAA±BBB = (a1, a2, a3)±(b1, b2, b3) = (a1±b1, a2±b2, a3±b3), for any vectors A
A
A and BBB in R3.
2. cAAA = c(a1, a2, a3) = (ca1, ca2, ca3), for any vector AAA and any number c.
Example 5. Let AAA = (1, −1, 0), BBB = (0, 1, 2). Compute the following vectors:
a. AAA + BBB = (1, −1, 0) + (0, 1, 2) = (1, 0, 2)
b. 2AAA = 2(1, −1, 0) = (2, −2, 0)
Having defined R2and R3, we now define Rn, i.e., the set of ordered n-tuples of real numbers.
Definition 2.3. Rn = {(x1, x2, . . . , xn) : x1, x2, . . . , xnare arbitrary real num- ber}. If AAA and BBB are any two vectors in Rn and a is any real number, we define vector addition and scalar multiplication in Rnas follows:
1. AAA ± BBB = (a1, a2, . . . , an) ± (b1, b2, . . . , bn) = (a1± b1, a2± b2, . . . , an± bn)
2.1. R2 THROUGH RN 63 2. cAAA = c(a1, a2, . . . , an) = (ca1, ca2, . . . , can)
We note that Rnmay also be thought of as the set of 1 ×n matrices. Sometimes we will want to think of Rn as the set of n × 1 matrices, i.e., Rn consists of columns rather than rows. One reason for this is so that Axxx will be defined for A an m × n matrix and xxx in Rn.
There are time when complex numbers are used for scalars. We then define Cn = {(z1, z2, . . . , zn) : zk are arbitrary complex numbesr}. Vector addition and scalar multiplication are defined as in Rn, except that the scalars may now be complex numbers instead of just real numbers.
Example 6. Write the vector (−1, 6, 4) as a sum of three vectors each of which is parallel to one of the coordinate axes.
(−1, 6, 4) = (−1, 0, 0) + (0, 6, 0) + (0, 0, 4) x3
x1
x2
(−1, 6, 0) (0, 6, 0)
(−1, 6, 4)
We can also write (−1, 6, 4) as a linear combination of the three vectors iii, jjj, and kkk.
(−1, 6, 4) = −(1, 0, 0) + 6(0, 1, 0) + 4(0, 0, 1) = −iii + 6jjj + 4kkk. A common mistake is to think that R2 is a subset of R3; that is, W = {(x1, x2, 0) : x1 and x2 arbitrary} is equated with R2. Clearly W is not R2, since W consists of triples of numbers, while R2 consists of pairs of numbers.
Example 7. Solve the following vector equation in R5: 2(−1, 4, 2, 0, 1) + 6XXX = 3(2, 0, 6, 1, −1)
6XXX = (6, 0, 18, 3, −3) − 2(−1, 4, 2, 0, 1) = (8, −8, 14, 3, −5) Thus,
X X X = 1
6(8, −8, 14, 3, −5) = 4 3, −4
3,7 3,1
2, −5 6
The algebraic operations we have defined in Rn have the same properties as those in R2, and we list them in the following theorem.
Theorem 2.2. Let AAA, BBB, and CCC be three arbitrary vectors in Rn. Let a and b be any two numbers. Then the following are true:
1. AAA + BBB = BBB + AAA
2. (AAA + BBB) + CCC = AAA + (BBB + CCC)
3. Let 000 = (0, 0, . . . , 0), then 000 + AAA = AAA
4. For every AAA in Rn there is a (−AAA) in Rn such that AAA + (−AAA) = 000. If A
A
A = (a1, . . . , an), −AAA = (−a1, . . . , −an) 5. a(AAA + BBB) = aAAA + aBBB
6. (a + b)AAA = aAAA + bAAA 7. (ab)AAA = a(bAAA) 8. 1AAA = AAA
We ask the reader to check that these identities are valid.
Problem Set 2.1
1. Sketch the following vectors in R2: a. (−1, 2), (0, 6), −2(1, 4)
b. (1, 1) + (−1, 1), 3(−2, 1) − 2(1, 1)
2. If AAA = (−1, 6) and BBB = (3, −2), sketch the vectors a. AAA, −AAA
b. AAA + BBB, −AAA + BBB, AAA − BBB c. 2AAA − 3BBB
3. Let AAA = (1, −1). Sketch all vectors of the form tAAA, where t is an arbitrary real number.
4. Let AAA = (1, −2). Let BBB = (−2, 6). Find XXX such that 6XXX + 2AAA = 3BBB.
5. Let AAA and BBB be the same vectors as in problem 4. Sketch vectors of the form XXX = c1AAA + c2BBB for various values of c1and c2. Which vectors in R2 can be written in this manner?
6. Prove Theorem 2.2 for n = 3.
7. Let AAA = (1, −1, 2), BBB = (0, 1, 0).
a. Find all vectors xxx in R3 such that xxx = t1AAA + t2BBB for some constants t1and t2.
b. Is the vector (0,0,1) one of the vectors you found in part a?
8. Let xxx = (1, 0), yyy = (0, 1). Give a geometrical description of the following sets of vectors.
2.2. DEFINITION OF A VECTOR SPACE 65 a. {txxx + (1 − t)yyy : 0 ≤ t ≤ 1}
b. {t1xxx + t2yyy : 0 ≤ t1≤ 1, 0 ≤ t2≤ 1}
c. {txxx + (1 − t)yyy : any real number}
d. {t1xxx + t2yyy : 0 ≤ t1, 1 ≤ t2} e. {t1xxx + t2yyy : t1 and t2 arbitrary}
9. Let xxx = (1, 2), yyy = (−1, 1). Describe the following sets of vectors.
a. {txxx + (1 − t)yyy : 0 ≤ t ≤ 1}
b. {t1xxx + t2yyy : 0 ≤ t1≤ 1, 0 ≤ t2≤ 1}
c. {txxx + (1 − t)yyy : t any real number}
d. {t1xxx + t2yyy : 0 ≤ t1, 1 ≤ t2} e. {t1xxx + t2yyy : t1 and t2 arbitrary}
10. Let xxx = (1, 1, 0) and let yyy = (1, 1, 1). Describe the following sets of vectors:
a. {txxx + (1 − t)yyy : 0 ≤ t ≤ 1}
b. {t1xxx + t2yyy : 0 ≤ t1≤ 1, 0 ≤ t2≤ 1}
c. {txxx + (1 − t)yyy: t any real nubmer}
11. Given two sets A and B, their intersection, A ∩ B, is defined as everything they have in common.
A ∩ B = {x: x ∈ A and x ∈ B}
Let
W1= {(x1, x2, x3) : x1= x2, x3= 0}
W2= {(x1, x2, x3) : x1= x2} W3= {(x1, x2, x3) : x1+ x2= 0}
Graph each of the following sets:
a. W1∩ W2 b. W2∩ W3 c. W1∩ W2∩ W3
2.2 Definition of a Vector Space
In the last section we defined the classical vector spaces Rn, for n = 2, 3, . . . . In this section we extract from these sets their crucial properties, and then define a vector space to be anything that has these properties. Thus, let V denote a set or collection of objects. The elements of V will be called vectors and we assume that there are two operations defined. The first (vector addition) will be denoted by a “+” sign, and the second (scalar multiplication) by a “·” or by juxtaposition. Specifically if xxx and yyy are any two vectors, i.e., elements of V , then xxx + yyy is in V . If a is any real number and xxx any vector, then axxx is
in V . The conditions that xxx + yyy and axxx be in V are expressed by saying that V is closed under vector addition and scalar multiplication. The term “in V ” has been underlined to underscore the idea that there may be times when an addition or multiplication has been defined such that the result will not be back in V . In such cases V is not a vector space.
In order for V to be a vector space, it is not enough that these two operations have been defined. They must also satisfy certain laws: those needed so that a reasonable arithmetic is possible. The requirements are listed in the following definition.
Definition 2.4. Let V be a set on which two operations, vector addition and scalar multiplication, have been defined. V will be called a real vector space if the following properties are true:
1. For any xxx and yyy in V, xxx + yyy is in V .
2. For any xxx in V , and any real number a in R, axxx is in V . 3. xxx + yyy = yyy + xxx, for any xxx, and zzz in V .
4. xxx + (yyy + zzz) = (xxx + yyy) + zzz, for any xxx, yyy, and zzz in V .
5. There is a vector in V denoted by 000 such that xxx + 000 = xxx, for every vector x
x x in V .
6. For any vector xxx in V , there is a vector (−xxx) in V such that xxx + (−xxx) = 000.
7. a(xxx + yyy) = axxx + ayyy, for any two vectors xxx and yyy in V and any real number a.
8. (a + b)xxx = axxx + bxxx, for any two real numbers a and b and any vector xxx.
9. (ab)xxx = a(bxxx), for any two real numbers a and b and any vector xxx.
10. 1xxx = xxx, for every vector xxx in V .
What the above properties mean practically is that all the algebraic manipu- lations you’ve learned to do with real numbers and vectors in Rn can also be done in this more abstract setting.
Remember, whenever someone says, “V is a real vector space” they are merely stating that the elements of V along with V ’s version of vector addition and scalar multiplication satisfy rules 1 through 10.
One more comment: The scalars are required to be real numbers and V is called a real vector space. There will be times when the scalars are complex numbers, in which case we call V a complex vector space. In the following we will not explicitly use the adjective real, since all our vector spaces will be real, unless the contrary is stated.
2.2. DEFINITION OF A VECTOR SPACE 67 Example 1. Rn(n ≥ 2) is a vector space. Definition 2.3 defined Rnas the set of n tuples of real numbers as well as what vector addition and scalar multiplication mean in this setting. Theorem 2.2 shows that 1 through 10 of Definition 2.4
hold.
Example 2. R1 = {x: x is a real number} with vector addition and scalar multiplication being ordinary addition and multiplication of real numbers. Example 3. Mmn, the set of all m × n matrices with real entries, with the usual way of adding matrices and multiplying matrices by numbers, is a vector
space.
Example 4. Let V = {(x1, 0, x3): such that x1 and x3 are real numbers}.
Define addition and scalar multiplication by xx
x + yyy = (x1, 0, x3) + (y1, 0, y3) = (x1+ y1, 0, x1+ y3) axxx = a(x1, 0, x3) = (ax1, 0, x3)!!
Clearly V is closed under these versions of vector addition and scalar multipli- cation; i.e., 1 and 2 hold. As a matter of fact, one can quickly check that 3 through 7 hold, but 8 does not. For
axxx + bxxx = a(x1, 0, x3) + b(x1, 0, x3) = (ax1, 0, x3) + (bx1, 0, x3)
= [(a + b)x1, 0, 2x3] while
(a + b)xxx = (a + b)(x1, 0, x3) = [(a + b)x1, 0, x3]
Since these two expressions will be equal only if 2x3= x3 or x3= 0 it is clear that 8 is not true for all the elements of V . Thus V is not a vector space. Example 5. Let V = {x: x > 0}. That is, V is the set of all positive real numbers. In this example ⊕ will denote vector addition and ⊙ will denote scalar multiplication. Define these operations as follows:
xx
x ⊕ yyy = xy; that is, vector addition will be ordinary multiplication a ⊙ xxx = xa
Since the product of two positive numbers is positive, and a positive number raised to any power is still positive, V is closed under these two operations; i.e., 1 and 2 in the definition of a vector space hold. We now proceed to check, as we must, each of the other required properties.
3. xxx ⊕ yyy = xy = yx = yyy ⊕ xxx
4. x ⊕ (yyy ⊕ zzz) = x(yz) = (xy)z = (xxx ⊕ yyy) ⊕ zzz
5. For the 000 vector we try the number 1 : 111 ⊕ xxx = 1x = x = xxx
6. For (−xxx) we try x−1: (−xxx) ⊕ xxx = x−1x = 1 = 000. Note that if x > 0, so is x−1.
7. a ⊙ (xxx ⊕ yyy) = (xy)a= (xa)(ya) = (a ⊙ xxx) ⊕ (a ⊙ yyy) 8. (a + b) ⊙ xxx = xa+b= xaxb= (a ⊙ xxx) ⊕ (b ⊙ xxx) 9. (ab) · xxx = xab= (xb)a= a ⊙ (b ⊙ xxx)
10. 1 ⊙ xxx = x1= xxx
Notice that in 5, the number 1 is treated as a vector in V , while in 10, it is treated as a scalar. Since V , with these definitions of vector addition and scalar multiplication, satisfies 1 through 10 of Definition 2.4, V is a vector space. The moral of this example, if there is one, is that neither “addition” nor
“multiplication” need always be what we’re used to. Notice that the zero vec- tor in the above example is not the number zero but the number 1. What is important about the zero vector is not that it be zero (the number) but that it satisfy the algebraic property 5 of Definition 2.4.
Example 6. V = {(x1, x2, 0) : x1, x2 are arbitrary real numbers}. The usual definitions of addition and multiplication of vectors in R3 will be used. The reader is asked to check that all 10 properties do indeed hold and conclude that V is a vector space.
Example 7. V = {fff : fff is a real-valued function defined on [0,1]}. For fff and ggg in V , fff + ggg must be in V ; that is, it must be a function defined on [0,1].
Thus for each t, 0 ≤ t ≤ 1, define (fff + ggg)(t) = fff(t) + ggg(t). We similarly define aggg, by (aggg)(t) = aggg(t). With these definitions of vector addition and scalar multiplication, 1 and 2 are satisfied. The reader is asked to check that the other
eight properties are also true.
Example 8. Let Pn= {polynomials in t of degree ≤ n} =
{antn+ an−1tn−1+ · · · + a1t + a0: where the aj are real numbers}
Thus
P0= {a: a is any real number} = R1
P1= {a0+ a1t : a0 and a1 are arbitrary real numbers}
If fff = f0+ · · · + fntn and ggg = g0+ · · · + gntn are any two polynomials in Pn, we define (fff + ggg)(t) by
[fff + ggg](t) = (f0+ g0) + (f1+ g1)t + · · · + (fn+ gn)tn Thus fff + ggg is in Pn. We define afff by
afff (t) = (af0) + (af1)t + · · · + (afn)tn
which is also in Pn. To see that Pnsatisfies the other conditions in Definition 2.4 is routine and is left to the reader. We note that Pn, for each value of n, is a
subset of the vector space in Example 7.
2.2. DEFINITION OF A VECTOR SPACE 69 We remind the reader that a vector space is not just a set, but a set with two operations: vector addition and scalar multiplication. If we change any one of these three, all ten properties must again be checked before we can say that we still have a vector space.
In the future when we deal with any of the standard vector spaces Rn, Pn, Mmn, or the vector space in Example 7, the operations of vector addition and scalar multiplication will not be explicitly stated. The reader, unless told otherwise, should assume the standard operations in these spaces.
Problem Set 2.2
1. Verify the details of Example 6.
2. Let V = {(x1, x2) : x1 and x2 are real numbers}. Define addition by (x1, x2) + (y1, y2) = (x1+ y1, x2− y2)
and scalar multiplication by
a(x1, x2) = (ax1, ax2)
Is V a vector space? If not, list all the axioms that V fails to satisfy.
3. Let V = {(x1, x2) : x1 and x2 are real numbers}. Define addition and scalar multiplication by
(x1, x2) + (y1, y2) = (x2+ y2, x1+ y1) a(x1, x2) = (ax1, ax2)
Is V a vector space? If not, list all the axioms that V fails to satisfy. What happens if we define a(x1, x2) = (ax2, ax1)?
4. Verify that Mmn(Example 3) is a vector space.
5. Verify that the set in Example 4 satisfies all the properties of Definition 2.4 except number 8.
6. Let C[0, 1] be the set of real-valued continuous functions defined on the interval [0,1]. This set is a subset of the vector space defined in Example 7.
Define addition and multiplication as in that example, and show that C[0, 1] is also a vector space.
7. Show that V in Example 7 is a vector space.
8. Show that Pn for n = 0, 1, . . . , is a vector space.
9. Let V0 be all 2 × 2 matrices for which the 1,1 entry is zero. Let V1 be all 2 × 2 matrices for which the 1,1 entry is nonzero. Determine if either of these two sets is a vector space. Define addition and multiplication as we did in Mmn.
10. Let V = {A, B, . . . , Z}; that is, V is the set of capital letters. Define A + B = C, B + C = D, A + E = F ; i.e., the sum of two letters is the first letter larger than either of the summands, unless Z is one of the letters to be added. In that case the sum will always be A. Can you define a way to multiply the elements in V by real numbers in such a way that V becomes a vector space?
11. Let V = {(x, 0, y): x and y are arbitrary real numbers}. Define addition and scalar multiplication as follows:
(x1, 0, y1) + (x2, 0, y2) = (x1+ x2, y1+ y2) c(x, 0, y) = (cx, cy)
Is V a vector space?
12. Let Vc= {(x1, x2) : x1+ 2x2= c}.
a. For each value of c sketch Vc.
b. For what values of c, if any, is Vc a vector space?
13. Let Vc =a1 a2 a3
a4 a5 a6
: a1+ a2+ a3= c
. For what values of c is Vc
a vector space?
14. Let Vc = {ppp: ppp is in P4and ppp(c) = 0}. For what values of c is Vc a vector space?
15. Let V = {A: A is in M23, and Σijaij = 0}. That is, A is in V if A is a 2 × 3 matrix whose entries sum to zero. Is V a vector space?
16. Let V1 = {(x1, x2) : x21+ x22 = 1}, V2 = {(x1, x2) : x21+ x22} ≤ 1, V3 = {(x1, x2) : x1≥ 0, x2≥ 0}, and V4 = {(x1, x2) : x1+ x2≥ 0}. Which of these subsets of R2 is a vector space?
17. Let V = {ppp: ppp is in P17 and ppp(1) + ppp(6) − ppp(2) = 0}. Is V a vector space? Does the subscript 17 have any bearing on the matter? What if the constraint is ppp(1) + ppp(6) − ppp(2) = 1?
2.3 Spanning Sets and Subspaces
The material covered so far has been relatively concrete and easy to absorb, but we now have to start thinking about an abstract concept, vector spaces. The only rules that we may use are those listed in Definition 2.4 and any others we are able to deduce from them. This, especially for the novice, is not easy and is somewhat tedious, but well worth the effort.
In the definition of a vector space, we have as axioms the existence of a zero vector (axiom 5) and an additive inverse (axiom 6) for every vector. One question that occurs is, how many zero vectors and inverses are there? Let’s see
2.3. SPANNING SETS AND SUBSPACES 71 if we can find an answer to this, at least in R2. Suppose kkk = (k1, k2) is another zero vector in R2. Thus xxx = xxx +kkk = (x1, x2) + (k1, k2) = (x1+ k1, x2+ k2). This implies that x1 = x1+ k1 and x2 = x2+ k2, from which k1 = k2 = 0. Hence kkk = (0, 0) = 000, which shows that R2 has a unique zero vector. Suppose now that yyy is an additive inverse for xxx in R2. Then 000 = xxx + yyy = (x1, x2) + (y1, y2) = (x1+ y1, x2+ y2). Hence x1+ y1= 0 = x2+ y2. Thus y1= −x1and y2= −x2, and we have uniqueness of inverses, at least in R2. What about other vector spaces?
Theorem 2.3. Let V be a vector space. Then
1. If 000 and kkk are both zero vectors (i.e., both satisfy axiom 5), then 000 = kkk.
2. If −xxx and yyy are both additive inverses of xxx, then −xxx = yyy.
Thus, in any vector space the zero vector is unique, and the additive inverse −xxx of any vector xxx is uniquely determined by xxx
Proof.
1. Since 000 satisfies 5, we have kkk + 000 = kkk but kkk also satisfies 5; thus 000 + kkk = 000.
These two equations plus commutativity of addition imply kkk = 000. Thus, there is only one zero vector.
2. yyy = yyy + 000 = yyy + [xxx + (−xxx)] = (yyy +xxx) + (−xxx) = 0 + (−xxx) = −xxx. We remind the reader that yyy + xxx = 000 since yyy is assumed to be an additive inverse for x
xx.
Some additional properties of vector spaces are listed in the following theorem.
The reader should try to prove them for V equal to R2; then read the proof of Theorem 2.4 and compare his or her version with ours.
Theorem 2.4. Let V be a vector space. Let XXX be any vector in V and a any number. Then
1. 0XXX = 000 2. a000 = 000 3. (−1)XXX = −XXX
4. aXXX = 000 only if a = 0 or XXX = 000 Proof.
1. Let xxx be any vector in V . Then, x
x
x + 0xxx = 1xxx + 0xxx = (1 + 0)xxx = 1xxx = xxx Adding −xxx to both sides of this equation, we get
000 = −xxx + xxx = −xxx + (xxx + 0xxx) = (−xxx + xxx) + 0xxx = 000 + 0xxx = 0xxx
2. If a = 0, then from 1 above we have a000 = 000. Thus we may suppose a 6= 0.
Then,
a000 + xxx = a000 + (aa−1)xxx = a(000 + a−1xxx) = a(a−1xxx) = 1xxx = xxx Hence from Theorem 2.3 a000 = 000.
3. xxx + (−1)xxx = 1xxx+(−1)xxx = [1+(−1)]xxx = 0xxx = 000. Theorem 2.3 now implies that (−1)xxx = −xxx.
4. Suppose axxx = 000 and a 6= 0. Then
000 = a−1000 = a−1(axxx) = (aa−1)xxx = 1xxx = xxx Thus axxx = 000 only when a = 0 or xxx = 000.
Quite often when the statement of a theorem or definition is read for the first time, its meaning is lost in a maze of strange words and concepts. The reader is strongly urged to always go back to R2or R3and try to understand what the statement means in this perhaps friendlier setting. For example, before proving Theorem 2.3 we analyzed the special case V = R2.
Given a collection of vectors {xxxk: k = 1, . . . , n}, we often wish to form arbitrary sums of these vectors; that is, we wish to look at all vectors xxx of the form
xx
x = c1xxx1+ c2xxx2+ · · · + cnxxxn=
n
X
k=1
ckxxxk (2.1) where the ck are arbitrary real numbers. Such sums are called linear combina- tions of the vectors xxxk.
Example 1. Let xxx1= (1, 2, 0), xxx2= (−1, 1, 0). Determine all linear combina- tions of these two vectors.
Solution. Any linear combination of these two vectors must be of the form xxx = c1xxx1+ c2xxx2
= c1(1, 2, 0) + c2(−1, 1, 0) = (c1, 2c1, 0) + (−c2, c2, 0)
= (c1− c2, 2c1+ c2, 0)
Thus, no matter what values the constants c1and c2 are, the third component of xxx will always be zero, and it seems likely that as c1 and c2vary over all pairs of real numbers, so too will the terms c1− c2and 2c1+ c2. We conjecture that the set of all linear combinations of these two vectors will be the set of vectors S in R3 whose third component is zero. To prove this, we only have to show that if xx is in S, then there are constants cx 1 and c2 such that xxx = c1xxx1+ c2xxx2. Thus suppose xxx is in S. Then xxx = (a, b, 0) for some numbers a and b. We want to find constants c1and c2 such that
(a, b, 0) = c1(1, 2, 0) + c2(−1, 1, 0) = (c1− c2, 2c1+ c2, 0)
2.3. SPANNING SETS AND SUBSPACES 73 or
c1− c2= a and 2c1+ c2= b The solution is
c1=a + b
3 c2=−2a + b 3
Thus (a, b, 0) is a linear combination of (1,2,0) and (−1, 1, 0). The set of all linear combinations of a set of vectors will occur frequently enough that we give it a special name and notation.
Definition 2.5. Given a set of vectors A = {xxxk: k = 1, 2, . . . , n} in some vector space V , the set of all linear combinations of these vectors will be called their linear span, or span, and denoted by S[A].
The result of the previous example restated in this notation is
S[(1, 2, 0), (−1, 1, 0)] = {(x1, x2, 0) : x1 and x2 are arbitrary real numbers}.
Example 2. Find the span of the vector (−1, 1).
(b) (a)
x2
x1
x1
x2
(−1, 1)
x1+ x2= 0
Figure 2.8
S[(−1, 1)] = {c(−1, 1): c is any real number}
= {(−c, c): c is any real number}
Thus the span of (−1, 1) may be thought of as the straight line x1+ x2= 0. Example 3. Find the span of A = {(1, 1, 0), (−1, 0, 0)}.
S[A] = {a(1, 1, 0) + b(−1, 0, 0): a and b arbitrary real numbers}
= {(a − b, a, 0): a, b arbitrary numbers}
= {(x, y, 0): x, y arbitrary numbers}
(a) (b) (1,1,0)
(−1, 0, 0) x2
x3 x3
x1
x1
x2
S[A] = x1, x2plane
Figure 2.9
The span of a nonempty set of vectors has some properties that we state and prove in the next theorem.
Theorem 2.5. Let S[A] be the linear span of the set A = {xxxk: k = 1, 2, . . . , n}.
Then S[A] satisfies axioms 1 through 10 of the definition of a vector space. That is, S[A] is a vector space.
Proof.
1. Let yyy and zzz be in S[A]. Then there are constants ajand bj, j = 1, 2, . . . , n such that
yyy =
n
X
j=1
ajxxxj zzz =
n
X
j=1
bjxxxj
Thus
yyy + zzz =
n
X
j=1
ajxxxj+
n
X
j=1
bjxxxj =
n
X
j=1
(ajxxxj+ bjxxxj)
=
n
X
j=1
(aj+ bj)xxxj
and we see that yyy + zzz is also in S[A].
2. Let xxx be in S[A] and k a real number. Then there are constants cj, j = 1, 2, . . . , n such that xxx =Pn
j=1cjxxxj. Thus
kxxx = k
n
X
j=1
cjxxxj
=
n
X
j=1
kcjxxxj
and we have that kxxx is in S[A].
3. and 4. Commutativity and associativity of vector addition are true be- cause we are in a vector space to begin with
2.3. SPANNING SETS AND SUBSPACES 75 5. Clearly each of the constants in a linear combination of the vectors xxxjcan
be set equal to zero. Thus we have
n
X
j=1
0xxxj =
n
X
j=1
000 = 000
and S[A] contains the zero vector.
6. Let xxx be in S[A]. Then xxx = Pn
j=1cjxxxj and −xxx = −(Pn
j=1cjxxxj) = Pn
j=1(−cj)xxxj. Thus, −xxx is also in S[A].
7. 8, 9, and 10 are true because they are true for all the vectors in V , and hence these properties are true for the vectors in S[A]. Remember every- thing in S[A] is automatically in V , since V is a vector space and is closed under linear combinations.
This fact, that the span of a set of vectors is itself a vector space contained in the original vector space, is expressed by saying that the span is a subspace of V .
Definition 2.6. Let V be a vector space. Let W be a nonempty subset of V . Then if W (using the same operations of vector addition and scalar multiplica- tion as in V ) is also a vector space, we say that W is a subspace of V .
Example 4. V = R3. Let W = {(r, 0, 0): r is any real number}. Show that W is a subspace of V .
x3
x2
W
x1
Figure 2.10
Solution. Since the operations of vector addition and scalar multiplication will be unchanged, we know that W satisfies properties 3, 4, 7, 8, 9, and 10. Hence, we merely need to verify that W satisfies 1, 2, 5, and 6 of Definition 2.4. Thus suppose that xxx and yyy are in W . Then xxx = (r, 0, 0) and yyy = (s, 0, 0) for some numbers r and s, and
xxx + yyy = (r, 0, 0) + (s, 0, 0) = (r + s, 0, 0)
Thus xxx + yyy is in W . Now let a be any real number, then axxx = a(r, 0, 0) = (ar, 0, 0) is also in W . To see that 000 and −xxx are in W we note that
000 = 0xxx = (0, 0, 0) and − xxx = −(r, 0, 0) = (−r, 0, 0)
Thus W is a subspace of V = R3.
Example 5. Let V be any vector space and let A be any nonempty subset of V . Then S[A] is a subspace of V . This is merely Theorem 2.5 restated. Example 6. Let xxx = (a, b, c) be any nonzero vector in R3. Then S[xxx], which is a subspace of R3, is just the straight line passing through the origin and the point with coordinates (a, b, c). See Figure 2.11. The reader should verify the
details of this example.
To verify that a subset of a vector space is a subspace can be tedious. The following theorem shows that it is sufficient to verify axioms 1 and 2 of Defini- tion 2.4.
xxx= (a, b, c)
Figure 2.11
Theorem 2.6. A nonempty subset W of a vector space V is a subspace of V if and only if
1. For any xxx and yyy in W , xxx + yyy is also in W .
2. For any scalar c and any vector xxx in W , cxxx is in W .
Proof. If W is a subspace of V , clearly 1 and 2 must be true. Thus, suppose 1 and 2 are true. Then we need to verify that axioms 3 through 10 in the definition of a vector space are also true. Since we have not changed how vectors are added or multiplied by scalars, axioms 3, 4, and 7 through 10 are automatically true.
Thus, only axioms 5 and 6 need to be verified.
2.3. SPANNING SETS AND SUBSPACES 77 5. To show that zero is in W , we use the fact that W is assumed to be nonempty. Let xxx be any vector in W . Then since W is closed under scalar multiplication we have 000 = 0xxx and 000 must be in W .
6. Let xxx be any vector in W , then (−1)xxx = −xxx must also be in W .
In the following, whenever we wish to indicate that a subspace W equals S[A], for some set A, we will say W is spanned by A or A is a spanning set of W .
Example 7. Let W be that subspace of R4spanned by the vectors (1, −1, 0, 2), (3,0,1,6), and (0, 1, 1, −1). Is the vector xxx = (2, −3, 4, 0) in W ?
Solution. If xxx is in W , it must be a linear combination of the three vectors whose span is W . Thus, there should exist constants c1, c2, and c3 such that
(2, −3, 4, 0) = c1(1, −1, 0, 2) + c2(3, 0, 1, 6) + c3(0, 1, 1, −1)
This vector equation has a solution if and only if the following system has a solution.
c1+ 3c2 = 2
−c1 + c3= −3 c2+ c3= 4 2c1+ 6c2− c3= 0
A few row operations on the augmented matrix of this system show that it is row equivalent to the matrix
1 3 0 2
0 1 0 0
0 0 1 4
0 0 0 −5
The nonhomogeneous system, which has the above matrix as its augmented matrix, has no solution. This means that W does not contain the given vector.
Example 8. Let V = R2. Let W = {(x, sin x): x is a real number}. Show that while W contains 000 and the additive inverse −xxx of every vector xxx in W , it still is not a subspace.
Solution. To see that 000 = (0, 0) is in W , we only need observe that sin 0 = 0.
Thus (0, sin 0) = (0, 0) is in W . If xxx is in W , then xxx = (r, sin r) for some number r. But then
−xxx = −(r, sin r) = (−r, − sin r) = (−r, sin(−r))
must also be in W . To see that W is not a subspace, we note that if W were a subspace, c(1, sin 1) must be in W for any choice of the scalar c. Since sin 1 > 0, by making c large enough we will have c sin 1 > 1. Since −1 ≤ sin x ≤ 1 for all x, we cannot have sin x = c sin 1 for any x. Thus, c(1, sin 1) = (c, csin 1) is not in W . Hence, W is not closed under scalar multiplication, and it cannot be a
subspace.
(1, sin (1))
c(1, sin (1))
4.0
1 π2 π
Figure 2.12
Example 9. Let A be an m × n matrix. Let K = (xxx : xxx is in Rn and Axxx = 000}.
Thus K is the solution set of the system of linear homogeneous equations whose coefficient matrix is A. Show that K is a subspace of Rn. Notice that Rn is being thought of as the set n × 1 matrices.
Solution. We first note that 000 is in K, since A000 = 000. Thus K is nonempty.
Suppose now that xxx1and xxx2 are in K. Then
A(xxx1+ xxx2) = Axxx1+ Axxx2= 000 + 000 = 000
and K is closed under addition. To see that K is closed under scalar multipli- cation we have, if xxx is in K and c is any number:
A(cxxx) = cA(xxx) = c000 = 000
Hence, by Theorem 2.6, K is a subspace of Rn.
Example 10. Let V = R4 and let W = {(x1, x2, x3, x4) : 2x1− x2+ x3= 0}.
Show W is a subspace of R4.
Solution. Note that W is the solution set of a system of homogeneous equations.
Thus, by the previous example W is a subspace.
Example 11. Show that F =1 −1
0 0
1 0 0 0
0 0 1 1
0 0 1 0
spans M22.
Solution. Let A =a1 a2
a3 a4
be an arbitrary vector (matrix) in M22. Then we have to find constants c1, c2, c3, and c4 such that
a1 a2
a3 a4
= c11 −1
0 0
+ c21 0 0 0
+ c30 0 1 1
+ c40 0 1 0
Thus, we have the equations
a1= c1+ c2 a2= −c1 a3= c3+ c4 a4= c3
2.3. SPANNING SETS AND SUBSPACES 79 Hence,
a1 a2
a3 a4
= (−a2)1 −1
0 0
+ (a1+ a2)1 0 0 0
+ a40 0 1 1
+ (a3− a4)0 0 1 0
Thus, M22= S[F ].
Problem Set 2.3
1. Show that S[(1, 0)(0, 1)] = R2.
2. Show that S[(1, 0, 0), (0, 1, 0)] equals the x1, x2plane in R3. 3. Show that S[(1, 0, 1)(1, 2, −1)] = {(x1, x2, x3) : x1− x2− x3= 0}.
4. Show that S[(1, 1)] = {(x1, x2) : x1− x2= 0}.
5. Let A = {(1, −1, 0), (−1, 0, 1)}. Determine which if any of the following vectors is in S[A] : (1, −1, 1), (0, −1, 1), (2, 3, −1).
6. Let A = {(1, 2), (6, 3)}. Show that S[A] = R2.
7. Let xxx = (x1, x2) be any nonzero vector in R2. Show that S[xxx] is a straight line passing through the origin and the point with coordinates (x1, x2).
8. What is S[(0, 0)]?
9. Let V be a vector space. Let W be that subset of V which consists of just the zero vector. Prove the following:
a. W is a subspace of V .
b. Show that V is a subspace of itself.
10. Let A be any set of vectors in some vector space V . Show that A is contained in S[A]. Show that if H is any subspace of V containing A, then S[A] is also in H. Thus, S[A] is the smallest subspace of V which contains A.
11. Let V = R3. Let A1 = {(1, 1, 0)}, A2 = {(1, 1, 0), (0, 0, 1)}. Show that S[A1] is contained in S[A2] and describe these two subspaces geometrically.
12. Let fff (x) = ax for some constant a. Show that the graph of fff = {(x, fff(x)):
x is any real number} is a subspace of R2. Conversely suppose we have a function fff (x) whose graph is a subspace of R2. Show that fff (x) = ax for some constant a. (Hint: What must a equal?)
13. Let A =1 0 3
4 4 6
. Let K be the solution set of Axxx = 000, for xxx in R3. From Example 9 we know that K is a subspace of R3. Find a vector xxx0
such that S[xxx0] = K.
14. Let W = {(x1, x2, x3) : x1− x2+ x3= c}. For which values of c, if any, is W a subspace of R3?
15. Example 11 exhibited a spanning set of M22 with four vectors. Can you find a spanning set of M22with five vectors? With three vectors?
16. Find spanning sets for each of the following vector spaces:
a. V = {(x1, x2, x3) : x1+ x2− 6x3= 0}
b. V = {ppp: ppp is in P2 and ppp(1) = 0}
c. V = {A = [aij] : A is in M23andP3
j=1aij= 0 for i = 1, 2}
17. Let V be the vector space of all n × n matrices. Show that the following subsets of V are subspaces:
a. W = all n × n scalar matrices = {cIn: c is any number}
b. W = all n × n diagonal matrices = {[djδjk}: dj any number}
c. W = all n × n upper triangular matrices d. W = all n × n lower triangular matrices
18. Let W be the set of all n × n invertible matrices. Show that W is not a subspace of the vector space of all n × n matrices.
19. Let W be any subspace of the vector space Mmn of all m × n matrices.
Show that WT, that subset of Mnm consisting of the transposes of all the matrices in W , is a subspace of Mnm.
20. Let P2 be all polynomials of degree 2 or less. Let W = {a + bt: a and b arbitrary real numbers}. Is W a subspace of P2?
21. Let V be a vector space. Let W1 and W2 be any two subspaces of V . Let W = W1∩ W2= {xxx : xxx belongs to both W1 and W2}. Show that W is a subspace of V .
22. Let V, W1, and W2 be as in problem 21. Let W = W1∪ W2 = {xxx: xxx belongs to W1 or W2}. Show that W need not be a subspace of V . 23. Let V, W1, and W2be as in problem 21. Define W1+ W2 as that subset of
V which consists of all possible sums of the vectors in W1and W2, that is W1+ W2= {uuu + vvv : uuu and vvv any vectors in W1and W2}
Show W1+ W2is a subspace of V and that any other subspace of V which contains W1 and W2 must also contain W1+ W2. Thus, W1+ W2 is the smallest subspace of V containing W1∪ W2.
2.4. LINEAR INDEPENDENCE 81 24. Let V = C[a, b]. Let W be that subset of V which consists of all polyno-
mials. Is W a subspace of V ?
25. Let W = {. . . , −2, −1, 0, 1, 2, . . .}. Is W a subspace of R1?
26. Let W = {(x1, x2) : x1 or x2 equals zero}. Let W1 = {(x1, 0)} and W2 = {(0, x2)}. Thus, W = W1∪ W2. Show that both W1 and W2 are subspaces of R2, and that W is not a subspace.
27. Let W = {(x1, x2) : x21+ x22≤ c}. For which values of c is W a subspace of R2?
28. Let W1 = {ppp: ppp is in P2 and ppp(1) = 0}. Let W2 = {ppp: ppp is in P2 and ppp(2) = 0}.
a. Find a spanning set for W1∩ W2.
b. Find spanning sets for W1 and W2 each of which contains the span- ning set you found in part a.
29. Example 9 shows that the solution sets of homogeneous systems of equa- tions are subspaces. Find spanning sets for each of the solution spaces of the following systems of equations:
a. 2x1+ 6x2− x3= 0
b. 2x1+ 6x2− x3= 0, x2+ 4x3= 0
c. 2x1+ 6x2− x3= 0, x2+ 4x3= 0, x1+ x2= 0 30. Let A =
2 6 1
−3 1 0
. Let V = {B : B ∈ M34 and AB = 024}.
a. Show V is a subspace of M34. b. Find a spanning set for V .
2.4 Linear Independence
In this section we discuss what it means to say that a set of vectors is linearly in- dependent. We encourage the reader to go over this material several times. The ideas discussed here are important, but hard to digest, thus needing rumination.
Consider the two vectors xxx = (1, 0, −1) and yyy = (0, 1, 1) in R3. They have the following property: Any vector zzz in their span can be written in only one way as a linear combination of these two vectors. Let’s check the details of this.
Suppose there are constants c1, c2, c3, and c4such that
zzz = c1xxx + c2yyy = c3xxx + c4yyy (2.2) What we want to show is that c1= c3 and c2= c4; that is, the coefficients of xxx must be equal and the coefficients of yyy must also be equal. The two expressions for zzz lead to the equation
000 = zzz − zzz = (c1− c3)xxx + (c2− c4)yyy (2.3)
Thus, we need to show that anytime 000 = axxx + byyy we must have a = 0 = b.
Suppose now that
000 = (0, 0, 0) = a(1, 0, −1) + b(0, 1, 1) = (a, b, −a + b)
Clearly this can occur only if a = 0 = b. This property of the two vectors xxx and yyy is given a special name.
Definition 2.7. A set of vectors {xxxk: k = 1, . . . , n} in a vector space V is said to be linearly independent if whenever
000 = c1xxx1+ c2xxx2+ · · · + cnxxxn=
n
X
k=1
ckxk
then
c1= c2= · · · = cn= 0
That is, the zero vector can be written in only one way as a linear combination of these vectors.
A set of vectors is said to be linearly dependent if it is not linearly indepen- dent. That is, a set of vectors {xxxk: k = 1, . . . , n} is linearly dependent only if there are constants c1, c2, . . . , cn not all zero such that 000 = c1xxx1+ c2xxx2+ · · · + cnxxxn.
Example 1. Show that the vectors {(1, 0, 1), (2, 0, 3), (0, 0, 1)} are linearly de- pendent.
Solution. We need to find three constants c1, c2, c3 not all zero such that c1(1, 0, 1) + c2(2, 0, 3) + c3(0, 0, 1) = (0, 0, 0)
Thus we have to find a nontrivial solution to the following system of equations:
c1+ 2c2 = 0 c1+ 3c2+ c3= 0
This system has many nontrivial solutions. A particular one is c1= 2, c2= −1, and c3 = 1. Since we have found a nontrivial linear combination of these vectors, namely, 2(1, 0, 1)−(2, 0, 3)+(0, 0, 1), which equals zero, they are linearly
dependent.
Example 2. Determine whether the set
{(1, 0, −1, 1, 2), (0, 1, 1, 2, 3), (1, 1, 0, 1, 0)}
is linearly dependent or independent.
Solution. We have to decide whether or not there are constants c1, c2, c3 not all zero such that
(0, 0, 0, 0, 0) = c1(1, 0, −1, 1, 2) + c2(0, 1, 1, 2, 3) + c3(1, 1, 0, 1, 0)
2.4. LINEAR INDEPENDENCE 83 The linear system that arises from this is
c1 + c3= 0 c2+ c3= 0
−c1+ c2 = 0 c1+ 2c2+ c3= 0 2c1+ 3c2 = 0
The only solution is the trivial one, c1 = c2 = c3 = 0. Thus the vectors are
linearly independent.
To say that a set of vectors is linearly independent is to say that the zero vector can be written in only one way as a linear combination of these vectors;
that is, every coefficient ck must equal zero. The following theorem shows that even more is true for a linearly independent set of vectors.
Theorem 2.7. If A = {xxxk: k = 1, . . . , p} is a linearly independent set of vectors, then every vector in S[A] can be written in only one way as a linear combination of the vectors in A.
Proof. Let yyy be any vector in S[A]. Suppose that we have
yyy =
p
X
k=1
akxxxk =
p
X
k=1
bkxxxk
We wish to show that ak= bk for each k. Subtracting yyy from itself we have 000 = yyy − yyy =
p
X
k=1
akxxxk−
p
X
k=1
bkxxxk=
p
X
k=1
(ak− bk)xxxk
Since the set A is linearly independent we must have
a1− b1= 0 a2− b2= 0 · · · ap− bp= 0
Thus, we can write any vector in S[A] as a linear combination of the linearly independent vectors xxxk in one and only one way.
The unique constants ak that are associated with any vector xxx in S[A] are given a special name.
Definition 2.8. Given a linearly independent set of vectors A = {xk: k = 1, . . . , n}. If xxx = Pn
k=1akxxxk, the coefficients ak, 1 ≤ k ≤ n, are called the coordinates of xxx with respect to A. Note that there is an explicit ordering of the vectors in the set A.
Example 3. Show that the set A = {(1, 1, −1), (0, 1, 0), (−1, 0, 2)} is linearly independent, and then find the coordinates of (2, −4, 6) with respect to A.
Solution. To see that A is linearly independent suppose c1(1, 1, −1) + c2(0, 1, 0) + c3(−1, 0, 2) = (0, 0, 0)
The system of equations that is equivalent to this vector equation, namely, c1 − c3= 0
c1+ c2 = 0
−c1 + 2c3= 0
has only the trivial solution. Thus A is a linearly independent set of vectors. We next want to write, if possible, (2, −4, 6) as a linear combination of the vectors in A. Thus we want constants ck, k = 1, 2, 3, such that
(2, −4, 6) = c1(1, 1, −1) + c2(0, 1, 0) + c3(−1, 0, 2) The associate system is
c1 − c3= 2 c1+ c2 = −4
−c1 + 2c3= 6
and its unique solution is c1 = 10, c2 = −14, c3 = 8. Thus (2, −4, 6) = 10(1, 1, −1) − 14(0, 1, 0) + 8(−1, 0, 2) and the coordinates of xxx = (2, −4, 6) with
respect to A are [10, −14, 8].
Example 4. Show that the set F is a linearly independent subset of M22. F =1 −1
0 0
1 0 0 0
0 0 1 1
0 0 1 0
Solution. Suppose there are constants c1, . . . , c4 such that
0 0 0 0
= c11 −1
0 0
+ c2
1 0 0 0
+ c3
0 0 1 1
+ c4
0 0 1 0
Thus, since the equations
c1+ c2= 0 − c1= 0 c3+ c4= 0 c3= 0
have only the trivial solution, F is linearly independent. From Example 11 in Section 2.3 we know that S[F ] = M22, and that
1 0 0 0
= 01 −1
0 0
+ 11 0 0 0
+ 00 0 1 1
+ 00 0 1 0
Thus, the coordinates of1 0 0 0
with respect to F are [0,1,0,0]. Example 5. Show that the set F = {1 + t, t2, t − 2} is a linearly independent subset of P2.
Solution. Suppose there are constants c1, c2, and c3 such that 0 = c1(1 + t) + c2(t2) + c3(t − 2)
= c2t2+ (c1+ c3)t + (c1− 2c3)
