vectors chapter 7. ch7_2 contents 7.1 vectors in 2-space 7.1 vectors in 2-space 7.2 vectors in...
TRANSCRIPT
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
Addition, Scalar Multiplication, Equality
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
yxadxdy
xadx
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
