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Vectors CHAPTER 7

# Vectors CHAPTER 7. Ch7_2 Contents 7.1 Vectors in 2-Space 7.1 Vectors in 2-Space 7.2 Vectors in 3-Space 7.2 Vectors in 3-Space 7.3 Dot Product 7.3

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Vectors

CHAPTER 7

Ch7_2

Contents

7.1 Vectors in 2-Space7.2 Vectors in 3-Space7.3 Dot Product7.4 Cross Product7.5 Lines and Planes in 3-Space7.6 Vector Spaces7.7 Gram-Schmidt Orthogonalization Process

Ch7_3

7.1 Vectors in 2-Space

Review of VectorsPlease refer to Fig 7.1 through Fig 7.6.

Ch7_4

Fig 7.1 (Geometric Vectors)

Ch7_5

Fig 7.2 (Vectors are equal)

Ch7_6

Fig 7.3 (Parallel vectors)

Ch7_7

Fig 7.4 (sum)

Ch7_8

Fig 7.5 (difference)

Ch7_9

Fig 7.6 (position vectors)

Ch7_10

Example 1

Please refer to Fig 7.7.Fig 7.7

Ch7_11

Let a = <a1, a2>, b = <b1, b2> be vectors in R2

(i) Addition: a + b = <a1 + a2, b1 + b2> (1)(ii) Scalar multiplication: ka = <ka1, ka2>,

k is a scalar (2)(iii)Equality: a = b if and only if a1 = b1, a2 = b2 (3)

DEFINITION 7.1

a – b = <a1− b1, a2 − b2>(4)

1 2 2 1 2 1 2 1,PP OP OP x x y y ������������������������������������������

Ch7_12

Graph Solution

Fig 7.8 shows the graph solutions of the addition and subtraction of two vectors.

Ch7_13

Example 2

If a = <1, 4>, b = <−6, 3>, find a + b, a − b, 2a + 3b.

Solution Using (1), (2), (4), we have

17,169,188,232

1,734),6(1

7,534),6(1

ba

ba

ba

Ch7_14

Properties

(i) a + b = b + a(ii) a + (b + c) = (a + b) + c(iii) a + 0 = a(iv) a + (−a) = 0(v) k(a + b) = ka + kb k scalar(vi) (k1 + k2)a = k1a + k2a k1, k2 scalars(vii) k1(k2a) = (k1k2)a k1, k2 scalars(viii) 1a = a(ix) 0a = 0 = <0, 0>

0 = <0, 0>

Ch7_15

Magnitude, Length, Norm

a = <a1 , a2>, then

Clearly, we have ||a|| 0, ||0|| = 0

22

21|||| aa a

Ch7_16

Unit Vector

A vector that ha magnitude 1 is called a unit vector. u = (1/||a||)a is a unit vector, since

1||||||||

1||||

1|||| a

aa

au

Ch7_17

Example 3

Given a = <2, −1>, then the unit vector in the same direction u is

and

51

,5

21,2

51

51 au

51

,5

2 u

Ch7_18

The i, j vectors

If a = <a1, a2>, then

(5)

Let i = <1, 0>, j = <0, 1>, then (5) becomes

a = a1i + a2j (6)

1,00,1,00,

,

2121

21

aaaa

aa

Ch7_19

Fig 7.10

Ch7_20

Example 4

(i) <4, 7> = 4i + 7j(ii) (2i – 5j) + (8i + 13j) = 10i + 8j(iii) (iv) 10(3i – j) = 30i – 10j(v) a = 6i + 4j, b = 9i + 6j are parallel

and b = (3/2)a

2|||| ji

Ch7_21

Example 5

Let a = 4i + 2j, b = –2i + 5j. Graph a + b, a – b

SolutionFig 7.11

Ch7_22

7.2 Vectors in 3-Space

Simple ReviewPlease refer to Fig 7.22 through Fig 7.24.

Fig 7.22

Ch7_23

Fig 7.23

Ch7_24

Fig 7.24

Ch7_25

Example 1

Graph the points (4, 5, 6) and (−2, −2, 0).Solution See Fig 7.25.

Ch7_26

Distance Formula

(1)

Fig 7.26

212

212

21221 )()()(),( zzyyxxPPd

Ch7_27

Example 2

Find the distance between (2, −3, 6) and (−1, −7, 4)

Solution

29)46())7(3())1(2( 222 d

Ch7_28

Midpoint Formula

(2)

2,

2,

2212121 zzyyxx

Ch7_29

Example 2

Find the midpoint of (2, −3, 6) and (−1, −7, 4)

SolutionFrom (2), we have

5 ,5 ,21

246

,2

)7(3,

2)1(2

Ch7_30

Vectors in 3-Space

Fig 7.27.

321 ,, aaaa

Ch7_31

Let a = <a1, a2 , a3>, b = <b1, b2, b3 > in R3

(i) a + b = <a1 + b1, a2 + b2, a3 + b3>(ii) ka = <ka1, ka2, ka3>(iii) a = b if and only if a1 = b1, a2 = b2, a3 = b3

(iv) –b = (−1)b = <− b1, − b2, − b3>(v) a – b = <a1 − b1, a2 − b2, a3 − b3>(vi) 0 = <0, 0 , 0>(vi)

DEFINITION 7.2

Component Definitions in 3-Spaces

23

22

21|||| aaa a

Ch7_32

Fig 7.28

Ch7_33

Example 4

Find the vector from (4, 6, −2) to (1, 8, 3)

Solution

5,2,3

)2(3,68,411221 OPOPPP

Ch7_34

Example 5

From Definition 7.2, we have

149

369476

73

72

||||222

a

Ch7_35

The i, j, k vectors

i = <1, 0, 0>, j = <0, 1, 0>, k = <0, 0, 1>

a = < a1, a2, a3> = a1i + a2j + a3j

1,0,00,1,00,0,1

,0,00,,00,0,

,,

321

321

321

aaa

aaa

aaa

Ch7_36

Fig 7.29

Ch7_37

Example 6a = <7, −5, 13> = 7i − 5j + 13j

Example 7(a) a = 5i + 3k is in the xz-plance(b)

Example 8If a = 3i − 4j + 8k, b = i − 4k, find 5a − 2b

Solution5a − 2b = 13i − 20j + 48k

3435||35|| 22 ki

Ch7_38

7.3 Dot Product

The dot product of a and b is the scalar

(1)

where is the angle between the vectors 0 .

DEFINITION 7.3Dot Product of Two Vectors

cos|||||||| baba ．

Ch7_39

Fig 7.32

Ch7_40

Example 1

From (1) we obtain

i i = 1, j j = 1, k k = 1(2)

Ch7_41

Component Form of Dot Product

(3)

(4)

See Fig 7.33

cos||||||||2|||||||||||| 22 baabc

222 ||||||||||(||2/1cos|||||||| cabba

332211 bababa ba．

Ch7_42

Fig 7.33

Ch7_43

Example 2

If a = 10i + 2j – 6k, b = (−1/2)i + 4j – 3k, then

21)3)(6()4)(2(21

)10(

ba．

Ch7_44

Properties

(i) a b = 0 if and only if a = 0 or b = 0(ii) a b = b a(iii) a (b + c) = a b + a c (iv) a (kb) = (ka) b = k(a b)(v) a a 0(vi) a a = ||a||2

Ch7_45

Orthogonal Vectors

(i) a b > 0 if and only if is acute(ii) a b < 0 if and only if is obtuse (iii) a b = 0 if and only if cos = 0, = /2

Note: Since 0 b = 0, we say the zero vector is orthogonal to every vector.

Two nonzero vectors a and b are orthogonal if and only if a b = 0.

THEOREM 7.1Criterion for an Orthogonal Vectors

Ch7_46

Example 3 i, j, k are orthogonal vectors.i j = j i = 0, j k = k j = 0, k i = i k = 0

(5)

Example 4If a = −3i − j + 4k, b = 2i + 14j + 5k, then

a b = –6 – 14 + 20 = 0They are orthogonal.

Ch7_47

Angle between Two Vectors

(6) ||||||||cos 332211

babababa

Ch7_48

Example 5

Find the angle between a = 2i + 3j + k, b = −i + 5j + k.

Solution

14,27||||,14|||| baba ．

942

271414

cos

44.9

77.0942

cos 1

Ch7_49

Direction Cosines

Referring to Fig 7.34, the angles , , are called the direction angles. Now by (6)

We say cos , cos , cos are direction cosines, and

cos2 + cos2 + cos2 = 1

||k||||a||ka

||j||||a||ja

||i||||a||ia ．．． cos,cos,cos

||a||||a||||a||321 cos,cos,cos

aaa

kjik||a||

j||a||

i||a||

a||a||

)(cos)(cos)(cos1 321 aaa

Ch7_50

Fig 7.34

Ch7_51

Example 6

Find the direction cosines and the direction angles of a = 2i + 5j + 4k.

Solution

5345452|||| 222 a

534

cos,53

5cos,

532

cos

Ch7_52

Component of a on b

Since a = a1i + a2j + a3k, then(7)

We write the components of a as(8)

See Fig 7.35. The component of a on any vector b is compba = ||a|| cos (9)

Rewrite (9) as

(10)

kajaia ．．． 321 ,, aaa

,comp iaai ． ,comp jaaj ． kaak ．comp

bba

bb

a

bba

bba

ab

||||1

||||||||cos||||||||

comp

Ch7_53

Fig 7.35

Ch7_54

Example 7

Let a = 2i + 3j – 4k, b = i + j + 2k. Find compba and compab.

SolutionForm (10), a b = −3

)2(6

1||||

1,6|||| kjib

bb

63

)2(6

1)432(comp kjikjiab ．

)432(291

||||1

,29|||| kjiaa

a

293

)432(291

)2(comp kjikjibb ．

Ch7_55

Physical Interpretation

See Fig 7.36. If F causes a displacement d of a body, then the work fone is

W = F d(11)

Ch7_56

Fig 7.36

Ch7_57

Example 8

Let F = 2i + 4j. If the block moves from (1, 1) to(4, 6), find the work done by F.

Solution d = 3i + 5j

W = F d = 26 N-m

Ch7_58

Projection of a onto b

See Fig 7.37. the projection of a onto i is

See Fig 7.38. the projection of a onto b is

(12)b

bbba

bb1

aa bb

)(compproj

iaiiaiaa ii 1)()(compproj

Ch7_59

Fig 7.37

Ch7_60

Fig 7.38

Ch7_61

Example 9

Find the projection of a = 4i + j onto b = 2i + 3j.

Solution

1311

)(2131

)(4comp 3jijiab

jijiab 1333

1322

)3(2131

1311

proj

Ch7_62

Fig 7.39

Ch7_63

7.4 Cross Product

The cross product of two vectors a and b is(1)

where is the angle between them, 0 , and nis a unit vector perpendicular to the plane of a and b with direction given by right-hand rule.

DEFINITION 7.4

Cross Product of Two Vectors

nbaba )sin||||||(||

Ch7_64

Fig 7.46

Ch7_65

Example 1

To understand the physical meaning of the cross product, please see Fig 7.37 and 7.48. The torque done by a force F acting at the end of position vector r is given by = r F.

Fig 7.47 Fig 7.48

Ch7_66

Properties

(i) a b = 0, if a = 0 or b = 0(ii) a b = −b a(iii) a (b + c) = (a b) + (a c)(iv) (a + b) c = (a c) + (b c)(v) a (kb) = (ka) b = k(a b)(vi) a a = 0(vii) a (a b) = 0(viii) b (a b) = 0

Two nonzero vectors a and b are parallel, if and only if a b = 0.

THEOREM 7.2Criterion for Parallel Vectors

Ch7_67

Example 2

(a) From properties (iv)i i = 0, j j = 0, k k = 0 (2)

(b) If a = 2i + 3j – k, b = –6i – 3j + 3k = –3a, then a and b are parallel. Thus a b = 0

If a = i, b = j, then

(3)

According to the right-hand rule, n = k. So i j = k

nnjiji

2sin||||||||

Ch7_68

Example 3

See Fig 7.49, we have

(4)(ii)property from and

jik

ikj

kji

jki

ijk

kii

Ch7_69

Fig 7.49

Ch7_70

Alternative Definition

Since

(5)

we have(6)

)()()(

)()()(

)()()(

)(

)()(

)()(

332313

322212

312111

3213

32123211

321321

kkjkik

kjjjij

kijiii

kjik

kjijkjii

kjikjiba

bababa

bababa

bababa

bbba

bbbabbba

bbbaaa

kjiba )()()( 122113312332 babababababa

Ch7_71

We also can write (6) as

(7)

In turn, (7) becomes

(8)

kjiba21

21

31

31

32

32

bb

aa

bb

aa

bb

aa

321

321

bbb

aaa

kji

ba

Ch7_72

Example 4

Let a = 4i – 2j + 5k, b = 3i + j – k, Find a b.

SolutionFrom (8), we have

kji

kji

ba

13

24

13

54

11

52

113

524

Ch7_73

Special Products

We have

(9)

is called the triple vector product. The following results are left as an exercise.

(10)

321

321

321

)(

ccc

bbb

aaa

cba．

cbabcacba )()()( ．．

Ch7_74

Area and Volume

Area of a parallelogram A = || a b||

(11)Area of a triangle

A = ½||a b||(12)Volume of the parallelepiped

V = |a (b c)|(13)See Fig 7.50 and Fig 7.51

Ch7_75

Fig 7.50

Ch7_76

Fig 7.51

Ch7_77

Example 5

Find the area of the triangle determined by the points (1, 1, 1), (2, 3, 4), (3, 0, –1).SolutionUsing (1, 1, 1) as the base point, we have two vectors a = <1, 2, 3>, b = <2, –1, –2>

kji

kji

kji

58

31

21

51

31

53

32

531

3213221

PPPP

1023

||58||21 kjiA

Ch7_78

Coplanar Vectors

a (b c) = 0 if and only if a, b, c are coplanar.

Ch7_79

7.5 Lines and Planes in 3-Space

Lines: Vector EquationSee Fig 7.55. We find r2 – r1 is parallel to r – r2, then

r – r2 = t(r2 – r1)(1)If we write

a = r2 – r1 = <x2 – x1, y2 – y1, z2 – z1> = <a1, a2, a3>

(2)then (1) implies a vector equation for the line is

r = r2 + tawhere a is called the direction vector.

Ch7_80

Fig 7.55

Ch7_81

Example 1

Find a vector equation for the line through (2, –1, 8) and (5, 6, –3).

SolutionDefine a = <2 – 5, –1 – 6, 8 – (– 3)> = <–3, –7, 11>.The following are three possible vector equations:

(3)

(4)

(5)

11,7,38,1,2,, tzyx

11,7,33,6,5,, tzyx

11,7,33,6,5,, tzyx

Ch7_82

Parametric equation

We can also write (2) as

(6)

The equations (6) are called parametric equations.

tazztayytaxx 322212 ,,

Ch7_83

Example 2

Find the parametric equations for the line in Example 1.

SolutionFrom (3), it follows

x = 2 – 3t, y = –1 – 7t, z = 8 + 11t (7)

From (5),

x = 5 + 3t, y = 6 + 7t, z = –3 – 11t (8)

Ch7_84

Example 3

Find a vector a that is parallel to the line: x = 4 + 9t, y = –14 + 5t, z = 1 – 3t

Solutiona = 9i + 5j – 3k

Ch7_85

Symmetric Equations

From (6)

provided ai are nonzero. Then

(9)

are said to be symmetric equation.

3

2

2

2

1

2

azz

ayy

axx

t

3

2

2

2

1

2

azz

ayy

axx

Ch7_86

Example 4

Find the symmetric equations for the line through (4, 10, −6) and (7, 9, 2)

SolutionDefine a1 = 7 – 4 = 3, a2 = 9 – 10 = –1, a3 = 2 – (–6) = 8, then

82

19

37

zyx

Ch7_87

Example 5

Find the symmetric equations for the line through (5, 3, 1) and (2, 1, 1)

SolutionDefine a1 = 5 – 2 = 3, a2 = 3 – 1 = 2, a3 = 1 – 1 = 0,then

1,2

33

5 z

yx

Ch7_88

Fig 7.56

Ch7_89

Example 6

Write vector, parametric and symmetric equations for the line through (4, 6, –3) and parallel to a = 5i – 10j + 2k.

SolutionVector: <x, y, z> = < 4, 6, –3> + t(5, –10, 2)

Parametric: x = 4 + 5t, y = 6 – 10t, z = –3 + 2t, Symmetric:

23

106

54

zyx

Ch7_90

Planes: Vector Equations

Fig 7.57(a) shows the concept of the normal vector to a plane. Any vector in the plane should be perpendicular to the normal vector, that is

n (r – r1) = 0 (10)

Ch7_91

Fig 7.57

Ch7_92

Cartesian Equations

If the normal vector is ai + bj + ck , then the Cartesian equation of the plane containing P1(x1, y1, z1) is

a(x – x1) + a(y – y1) + c(z – z1) = 0(11)

Ch7_93

Example 7

Find the plane contains (4, −1, 3) and is perpendicular to n = 2i + 8j − 5k

SolutionFrom (11):

2(x – 4) + 8(y + 1) – 5(z – 3) = 0or

2x + 8y – 5z + 15 = 0

Ch7_94

Equation (11) can always be written as ax + by + cz + d = 0

(12)

The graph of any ax + by + cz + d = 0, a, b, c not all

zero, is a plane with the normal vector n = ai + bj + ck

THEOREM 7.3Plane with Normal Vector

Ch7_95

Example 8

A vector normal to the plane 3x – 4y + 10z – 8 = 0 is n = 3i – 4j + 10k.

Ch7_96

Given three noncollinear points, P1, P2, P3, we arbitrarily choose P1 as the base point. See Fig 7.58, Then we can obtain

(13)0)()]()[( 11312 rrrrrr ．

Ch7_97

Fig 7.58

Ch7_98

Example 9

Find an equation of the plane contains (1, 0 −1), (3, 1, 4) and (2, −2, 0).SolutionWe arbitrarily construct two vectors from these three points, say, u = <2, 1, 5> and v = <1, 3, 4>.

,)2()2(),,(

)0,2,2(

,43)0,2,2(

)4,1,3(

,52)4,1,3(

)1,0,1(

kjiw

kjiv

kjiu

zyxzyx

Ch7_99

Example 9 (2)

If we choose (2, −2, 0) as the base point, then<x – 2, y + 2, z – 0> <−11, −3, 5> = 0

kji

kji

vu 5311

431

512

05)2(3)2(11 zyx

0165311 zyx

Ch7_100

Graphs

The graph of (12) with one or two variables missing is still a plane.

Ch7_101

Example 10

Graph 2x + 3y + 6z = 18

SolutionSetting: y = z = 0 gives x = 9

x = z = 0 gives y = 6x = y = 0 gives z = 3

See Fig 7.59.

Ch7_102

Fig 7.59

Ch7_103

Example 11

Graph 6x + 4y = 12

SolutionThis equation misses the variable z, so the plane is parallel to the z-axis. Since x = 0 gives y = 3

y = 0 gives x = 2See Fig 7.60.

Ch7_104

Fig 7.60

Ch7_105

Example 12

Graph x + y – z = 0

SolutionFirst we observe that the plane passes through (0, 0, 0). Let y = 0, then z = x; x = 0, then z = y.

Ch7_106

Fig 7.61

Ch7_107

Two planes that are not parallel must intersect in a line. See Fig 7.62. Fig 7.63 shows the intersection of a line and a plane.

Ch7_108

Fig 7.62

Ch7_109

Fig 7.63

Ch7_110

Example 13

Find the parametric equation of the line of the intersection of

2x – 3y + 4z = 1 x – y – z = 5

SolutionFirst we let z = t,

2x – 3y = 1 – 4t x – y = 5 + tthen x = 14 + 7t, y = 9 + 6t, z = t.

Ch7_111

Example 14

Find the point of intersection of the plane 3x – 2y + z = −5 and the line x = 1 + t, y = −2 + 2t, z = 4t.

SolutionAssume (x0, y0, z0) is the intersection point.

3x0 – 2y0 + z0 = −5 and x0 = 1 + t0, y0 = −2 + 2t0, z0 = 4t0

then 3(1 + t0) – 2(−2 + 2t0) + 4t0 = −5, t0 = −4Thus, (x0, y0, z0) = (−3, −10, −16)

Ch7_112

7.6 Vector Spaces

n-SpaceSimilar to 3-space

(1)

(2)

nn bababa ,,, 2211 ba nkakakak ,,, 21 a

nn

nn

bababa

bbbaaa

2211

2121 ,,,,,, ．．ba

Ch7_113

Let V be a set of elements on which two operations, vector addition and scalar multiplication, are defined. Then V is said to be a vector spaces if the following are satisfied.

DEFINITION 7.5

Vector Space

Ch7_114

Axioms for Vector Addition(i) If x and y are in V, then x + y is in V.(ii) For all x, y in V, x + y = y + x(iii) For all x, y, z in V, x + (y + z) = (x + y) + z(iv) There is a unique vector 0 in V, such that

0 + x = x + 0 = x(v) For each x in V, there exists a vector −x in V,

such that x + (−x) = (−x) + x = 0

DEFINITION 7.5

Vector Space

Ch7_115

Axioms for Scalars Multiplication

(vi) If k is any scalar and x is in V, then kx is in V.

(vii) k(x + y) = kx + ky

(viii) (k1+k2)x = k1x+ k2x

(ix) k1(k2x) = (k1k2)x

(x) 1x = x

Properties (i) and (vi) are called the closure axioms.

DEFINITION 7.5

Vector Space

Ch7_116

Example 1

Determine whether the sets (a) V = {1} and (b) V = {0} under ordinary addition and multiplication by real numbers are vectors spaces.

Solution (a) V = {1}, violates many of the axioms.(b) V = {0}, it is easy to check this is a vector space.

Moreover, it is called the trivial or zero vectorspace.

Ch7_117

Example 2

Consider the set V of all positive real numbers. If x and y denote positive real numbers, then we write vectors as x = x, y = y. Now addition of vectors is defined by

x + y = xyand scalar multiplication is defined by

kx = xk

Determine whether the set is a vector space.

Ch7_118

Example 2 (2)

Solution We go through all 10 axioms.(i) For x = x > 0, y = y > 0 in V, x + y = x + y > 0

(ii) For all x = x, y = y in V, x + y = x + y = y + x = y + x

(iii) For all x = x , y = y, z = z in Vx + (y + z) = x(yz) = (xy) = (x + y) + z

(iv) Since 1 + x = 1x = x = x, x + 1 = x1 = x = xThe zero vector 0 is 1 = 1

Ch7_119

Example 2 (3)

(v) If we define −x = 1/x, thenx + (−x) = x(1/x) = 1 = 1 = 0−x + x = (1/x)x = 1 = 1 = 0

(vi) If k is any scalar and x = x > 0 is in V, then kx = xk > 0

(vii) If k is any scalar, k(x + y) = (xy)k = xkyk = kx + ky

(viii) (k1+k2)x = xk1+k2 = xk1xk2 = k1x+ k2x

(ix) k1(k2x) = (xk2 )k1 = xk1k2 = (k1k2)x

(x) 1x = x1 = x = x

Ch7_120

If a subset W of a vector space V is itself a vector space under the operations of vector addition and scalar multiplication defined on V, then W is called a subspace of V.

DEFINITION 7.6

Subspace

Ch7_121

A nonempty subset W is a subspace of V if and only if

W is closed under vector addition and scalar

multiplication defined on V:(i) If x and y are in W, then x + y is in W.(ii) If x is in W and k is any scalar, then kx is in W.

THEOREM 7.4Criteria for a Subspace

Ch7_122

Example 3

Suppose f and g are continuous real-valued functions defined on (−, ). We know f + g and kf, for any real number k, are continuous real-valued. From this, we conclude that C(−, ) is a subspace of the vector space of real-valued function defined on (−, ).

Ch7_123

Example 4

The set Pn of polynomials of degree less than or equal to n is a subspace of C(−, ).

Ch7_124

A set of vectors {x1, x2, …, xn} is said to be linearly

independent, if the only constants satisfying k1x1 + k2x2 + …+ knxn = 0 (3)

are k1= k2 = … = kn = 0. If the set of vectors is not

linearly independent, it is linearly dependent.

DEFINITION 7.7

Linear Independence

Ch7_125

For example: i, j, k are linearly independent.

<1, 1, 1> , <2, –1, 4> and <5, 2, 7> are linearly dependent, because

3<1, 1, 1> + <2, –1, 4> − <5, 2, 7> = <0, 0, 0>3a + b – c = 0

Ch7_126

It can be shown that any set of three linearly independent vectors is a basis for R3. For example

<1, 0, 0>, <1, 1, 0>, <1, 1, 1>

Consider a set of vectors B = {x1, x2, …, xn} in a vector space V. If the set is linearly independent and if every vector in V can be expressed as a linear combination of these vectors, then B is said to be a basis for V.

DEFINITION 7.8

Basis for a Vector Space

Ch7_127

Standard Basis: {i, j, k} For Rn : e1 = <1, 0, …, 0>, e2 = <0, 2, …, 0> …..

en = <0, 0, …, 1>(4)If B is a basis, then there exists scalars such that

(5)

where these scalars ci, i = 1, 2, .., n, are called the coordinates of v related to the basis B.

cnccc xxxv 2211

Ch7_128

The number of vectors in a basis B for vector space V

is said to be the dimension of the space.

DEFINITION 7.8Dimension for a Vector Space

Ch7_129

Example 5

(a) The dimensions of R, R2, R3, Rn are in turn 1, 2, 3, n.

(b) There are n + 1 vectors in B = {1, x, x2, …, xn}. The dimension is n + 1

(c) The dimension of the zero space {0} is zero.

Ch7_130

Linear DEs

The general solution of following DE

(6)

can be written as y = c1y1 + c1y1 + … cnyn and it is said to be the solution space. Thus {y1, y2, …, yn} is a basis.

0)()()()( 011

1

1

ydxa

dx

ydxa n

n

nn

n

n

Ch7_131

Example 6

The general solution of y” + 25y = 0 is

y = c1 cos 5x + c2 sin 5x

then {cos 5x , sin 5x} is a basis.

Ch7_132

Span

If S denotes any set of vectors {x1, x2, …, xn} then the linear combination

k1x1 + k2x2 + … + knxn

is called a span of the vectors and written as Span(S) or Span{x1, x2, …, xn}.

Ch7_133

Rephrase Definition 7.8 and 7.9

A set S of vectors {x1, x2, …, xn} in a vector space V is a basis, if S is linearly independent and is a spanning set for V. The number of vectors in this spanning set S is the dimension of the space V.

Ch7_134

7.7 Gram-Schmidt Orthogonalization Process

Orthonormal Basis All the vectors of a basis are mutually orthogonal and have unit length .

Ch7_135

Example 1

The set of vectors

(1)

is linearly independent in R3. Hence B = {w1, w2, w3} is a basis. Since ||wi|| = 1, i = 1, 2, 3, wi wj = 0, i j, B is an orthonormal basis.

21

,2

1 0,

,6

1 ,

61

,6

2

,3

1 ,

31

,3

1

3

2

1

w

w

w

Ch7_136

Proof Since B = {w1, w2, …, wn} is an orthonormal basis, then any vector can be expressed as

u = k1w1 + k2w2 + … + knwn (2)(u wi) = (k1w1 + k2w2 + … + knwn) wi

= ki(wi wi) = ki

Suppose B = {w1, w2, …, wn} is an orthonormal basis

for Rn, If u is any vector in Rn, then

u = (u w1)w1 + (u w2)w2 + … + (u wn)wn

THEOREM 7.5

Coordinates Relative ti an Orthonormal Basis

Ch7_137

Example 2

Find the coordinate of u = <3, – 2, 9> relative to the orthonormal basis in Example 1.

Solution

321

321

211

61

310

211

,6

1 ,

310

wwwu

wuwuwu

Ch7_138

Gram-Schmidt Orthogonalization Process

The transformation of a basis B = {u1, u2} into an orthogonal basis B’= {v1, v2} consists of two steps. See Fig 7.64.

(3)1

11

1222

11

vvvvu

uv

uv

Ch7_139

Fig 7.64(a)

Ch7_140

Fig 7.64(b)

Ch7_141

Fig 7.64(c)

Ch7_142

Example 3

Let u1 = <3, 1>, u2 = <1, 1>. Transform them into an orthonormal basis.

Solution From (3)

Normalizing:

See Fig 7.65

53 ,

51

1 3,104

1 1,

1 3,

2

11

v

uv

103

,1011

101

,1031

22

2

11

1

vv

w

vv

w

Ch7_143

Fig 7.65

Ch7_144

Gram-Schmidt Orthogonalization Process

For R3:

(4)

222

231

11

1333

111

1222

11

vvvvu

vvvvu

uv

vvvvu

uv

uv

Ch7_145

See Fig 7.66. Suppose W2 = Span{v1, v2}, then

is in W2 and is called the orthogonal projection of u3 onto W2, denoted by x = projw2u3.

(5)

(6)

222

231

11

1332

proj vvvvu

vvvvu

ux

w

111

1221

proj vvvvu

ux

w

222

231

11

13 vvvvu

vvvvu

x

Ch7_146

Fig 7.66

Ch7_147

Example 4

Let u1 = <1, 1, 1>, u2 = <1, 2, 2>, u3 = <1, 1, 0>. Transform them into an orthonormal basis.Solution From (4)

21

,21

0,

31 ,

31 ,

32

1) 1, 1,35

2 2, 1,

1 1, 1,

222

231

11

1333

111

1222

11

vvvvu

vvvvu

uv

vvvvu

uv

uv

Ch7_148

Example 4 (2)

,2

1 ,

21

0,

,6

1 ,

61

,6

2 ,

31

,3

1 ,

31

, ,

3, 2, 1, ,1

and 22

,36

,3

21

,21

,0 ,31 ,

31 ,

32

,1 1, 1, v, v,v

3

21

321

321

321

w

ww

www

vv

wvvv

B

i

B

ii

i

Ch7_149

Let B = {u1, u2, …, um}, m n, be a basis for a Subspace Wm of

Rn. Then {v1, v2, …, vm}, where

is an orthogonal basis for Wm. An orthonormal basis for Wm is

THEOREM 7.6Orthogonalization Process

111

12

22

21

11

1

222

231

11

1333

111

1222

11

mmm

mmmmmm v

vvvu

vvvvu

vvvvu

uv

vvvvu

vvvvu

uv

vvvvu

uv

uv

mm

mB vv

vv

vv

www1

, ,1

,1

, , , 22

11

21