chapter 4 vector spaces linear algebra. ch04_2 definition 1. let be a sequence of n real numbers....
TRANSCRIPT
Chapter 4
Vector Spaces
Linear Algebra
Ch04_2
Definition 1.Let be a sequence of n real numbers. The set of all such sequences is called n-space (or n-dimensional space) and is denoted Rn.
u1 is the first component of .
u2 is the second component and so on.
) ..., , ,( 21 nuuu
) ..., , ,( 21 nuuu
Example 1
• R2 is the collection of all sets of two ordered real numbers. For example, (0, 0) , (1, 2) and (-2, -3) are elements of R2.
• R3 is the collection of all sets of three ordered real numbers.
For example, (0,0, 0) and (-1,3, 4) are elements of R3.
4.1 The vector Space Rn
Ch04_3
Definition 2. Let be two elements of Rn.
We say that u and v are equal if u1 = v1, …, un = vn.
Thus two elements of Rn are equal if their corresponding components are equal.
) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu
Definition 3. Let be elements of Rn
and let c be a scalar.
Addition and scalar multiplication are performed as follows:
Addition:
Scalar multiplication :
) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu
) ..., ,() ..., ,(
1
11
n
nn
cucucvuvu
uvu
Ch04_4
► The set Rn with operations of componentwise addition and scalar multiplication is an example of a vector space, and its elements are called vectors. We shall Henceforth interpret Rn to be a vector space.(We say that Rn is closed under addition and scalar multiplication).
► In general, if u and v are vectors in the same vector space, then u + v is the diagonal of the parallelogram defined by u and v.
Figure 4.2
Ch04_5
Example 2
Let u = ( –1, 4, 3) and v = ( –2, –3, 1) be elements of R3.
Find u + v and 3u.
Solution: u + v = (–1, 4, 3) + (– 2, –3, 1) = (-3 ,1 ,4)
3u = 3 (–1, 4, 3) = (-3 ,12 ,9)Example 3
Figure 4.1
In R2 , consider the two elements (4, 1) and (2, 3).Find their sum and give a geometrical
interpretation of this sum.
we get (4, 1) + (2, 3) = (6, 4).
The vector (6, 4), the sum, is the diagonal of the parallelogram.
Ch04_6
Example 4
Figure 4.3
Consider the scalar multiple of the vector (3, 2) by 2, we get
2(3, 2) = (6, 4)
Observe in Figure 4.3 that (6, 4) is a vector in the same direction as (3, 2), and 2 times it in length.
Ch04_7
Zero Vector The vector (0, 0, …, 0), having n zero components, is called the zero vector of Rn and is denoted 0.
Negative Vector
The vector (–1)u is writing –u and is called the negative of u. It is a vector having the same length (or magnitude) as u, but lies in the opposite direction to u.
Subtraction
Subtraction is performed on element of Rn by subtracting corresponding components.
u
u
Ch04_8
Theorem 4.1Let u, v, and w be vectors in Rn and let c and d be scalars.
(a) u + v = v + u
(b) u + (v + w) = (u + v) + w
(c) u + 0 = 0 + u = u
(d) u + (–u) = 0
(e) c(u + v) = cu + cv
(f) (c + d)u = cu + du
(g) c(du) = (cd)u
(h) 1u = u
Figure 4.4 Commutativity of
vector addition u + v = v + u
Ch04_9
Example 5 Let u = (2, 5, –3), v = ( –4, 1, 9), w = (4, 0, 2) in the vector space R3. Determine the vector 2u – 3v + w.
Solution
)31 7, 20,(
)2276 ,0310 4,12(4
2) 0, (4,27) 3, ,12()6 10, ,4(
2) 0, (4,9) 1, ,4(3)3 5, ,2(2
w3v2u
Ch04_10
Column Vectors
nnnn vu
vu
v
v
u
u
1111
We defined additionaddition and scalar multiplicationscalar multiplication of column vectors in Rn in a componentwise manner:
and
nn cu
cu
u
uc
11
Row vector:
Column vector:
) ..., , ,( 21 nuuuu
nu
u
1
Ch04_11
Homework
Exercise set 1.3 page 32:3, 5, 7, 9.
Ch04_12
4.2 Dot Product, Norm, Angle, and Distance
Definition Let be two vectors in Rn. The dot product of u and v is denoted u.v and is defined by .
The dot product assigns a real number to each pair of vectors.
) ..., , ,( and ) ..., , ,( 2121 nn vvvuuu vu
nnvuvu 11vu
Example 1
Find the dot product ofu = (1, –2, 4) and v = (3, 0, 2)
Solution
11
803
)24()02()31(
vu
Ch04_13
Properties of the Dot Product
Let u, v, and w be vectors in Rn and let c be a scalar. Then
1. u.v = v.u
2. (u + v).w = u.w + v.w
3. cu.v = c(u.v) = u.cv
4. u.u 0, and u.u = 0 if and only if u = 0
Proof1.
uv
vuvu
numbers real ofproperty ecommutativ by the
get We). ..., , ,( and ) ..., , ,(Let
11
11
2121
nn
nn
nn
uvuvvuvu
vvvuuu
4.
. ifonly and if 0 Thus.0 , ,0 ifonly and if ,0
.0 thus,0
122
1
221
22111
0uuu
uu
uu
nn
n
nnn
uuuu
uu
uuuuuu
Ch04_14
Norm of a Vector in Rn
Definition The norm (length or magnitude) of a vector u = (u1, …, un) in Rn
is denoted ||u|| and defined by
Note: The norm of a vector can also be written in terms of the dot product
221 nuu u
uuu
Figure 4.5 length of u
Ch04_15
Definition A unit vector is a vector whose norm is 1.
If v is a nonzero vector, then the vector
is a unit vector in the direction of v.
This procedure of constructing a unit vector in the same direction as a given vector is called normalizing the vector.
vv
u1
Find the norm of each of the vectors u = (1, 3, 5) of R3 and v = (3, 0, 1, 4) of R4.
Solution 352591)5()3()1( 222 u
2616109)4()1()0()3( 2222 v
Example 2
Ch04_16
Example 3
Solution
(a) Show that the vector (1, 0) is a unit vector.
(b) Find the norm of the vector (2, –1, 3). Normalize this vector.
(a) Thus (1, 0) is a unit vector. It can be similarly shown that (0, 1) is a unit vector in R2.
.1010) ,1( 22
(b) The norm of (2, –1, 3) is
The normalized vector is
The vector may also be written
This vector is a unit vector in the direction of (2, –1, 3).
.143)1(23) ,1 ,2( 222 .14
)3 ,1 ,2(141
.143
,141
,142
Ch04_17
Angle between Vectors ( in R2)
← Figure 4.6
The law of cosines gives:
Ch04_18
Definition Let u and v be two nonzero vectors in Rn.
The cosine of the angle between these vectors is
0 cosvuvu
Example 4Determine the angle between the vectors u = (1, 0, 0) and v = (1, 0, 1) in R3.
Solution 11) 0, (1,0) 0, 1,( vu
2101 1001 222222 vu
Thus the angle between u and v is 45.,2
1cos
vuvu
Angle between Vectors (in R n)
Ch04_19
Definition Two nonzero vectors are orthogonal if the angle between them is a
right angle .
Two nonzero vectors u and v are orthogonal if and only if u.v = 0.
Theorem 4.2
Proof
0 0cos orthogonal are , vuvu
Orthogonal Vectors
Ch04_20
Solution
Example 5Show that the following pairs of vectors are orthogonal.
(a) (1, 0) and (0, 1).
(b) (2, –3, 1) and (1, 2, 4).
(a) (1, 0).(0, 1) = (1 0) + (0 1) = 0.
The vectors are orthogonal.
(b) (2, –3, 1).(1, 2, 4) = (2 1) + (–3 2) + (1 4) = 2 – 6 + 4 = 0.
The vectors are orthogonal.
Ch04_21
Note
(1, 0), (0,1) are orthogonal unit vectors in R2.
(1, 0, 0), (0, 1, 0), (0, 0, 1) are orthogonal unit vectors in R3.
(1, 0, …, 0), (0, 1, 0, …, 0), …, (0, …, 0, 1) are orthogonal unit vectors in Rn.
Ch04_22
Example 6
Determine a vector in R2 that is orthogonal to (3, –1). Show that there are many such vectors and that they all lie on a line.
Solution
Let the vector (a, b) be orthogonal to (3, –1). We get
abba
baba
3030))1(()3(01) ,3() ,(
Thus any vector of the form (a, 3a)is orthogonal to the vector (3, –1). Any vector of this form can be written
a(1, 3)The set of all such vectors lie on the line defined by the vector (1, 3).
Figure 4.7
Ch04_23
Theorem 4.3
Let u and v be vectors in Rn.
(a) Triangle Inequality:
||u + v|| ||u|| + ||v||.
(a) Pythagorean theorem : If u.v = 0 then ||u + v||2 = ||u||2 + ||v||2.
Figure 4.8(a) Figure 4.8(b)
Ch04_24
Distance between PointsLet be two points in Rn.
The distance between x and y is denoted d(x, y) and is defined by
Note: We can also write this distance as follows.
) ..., , ,( and ) ..., , ,( 2121 nn yyyxxx yx
2211 )()() ,( nn yxyxd yx
yxyx ) ,(d
Example 7. Determine the distance between the points x = (1,–2 , 3, 0) and y = (4, 0, –3, 5) in R4.
Solution
74
253649
)50()33()02()41(),( 2222
yxd
x
yxy
Note: ( , ) ( , ) ( )It is clear that d d the symmetric propertyx y y x
Ch04_25
Homework
Exercise set 1.5 pages 47 to 48:2, 6, 10, 16, 20, 33, 35.
Exercise 36
Let u and v be vectors in Rn. Prove that ||u|| = ||v|| if and only if u + v and u v are orthogonal.
Ch04_26
4.3 General Vector Spaces
Definition A vector space is a set V of elements called vectors, having operations of addition and scalar multiplication defined on it that satisfy the following conditions.
Let u, v, and w be arbitrary elements of V, and c and d are scalars.
• Closure Axioms
1. The sum u + v exists and is an element of V. (V is closed under addition.)
2. cu is an element of V. (V is closed under scalar multiplication.)
Our aim in this section will be to focus on the algebraic properties of Rn.
Ch04_27
Example 1
(1) V={ …, 3, 1, 1, 3, 5, 7, …} V is not closed under addition because 1+3=4 V.
(2) Z={ …, 2, 1, 0, 1, 2, 3, 4, …} Z is closed under addition because for any a, b Z, a + b Z. Z is not closed under scalar multiplication because ½ is a scalar, for any odd a Z, (½)a Z.
Ch04_28
• Addition Axioms
3. u + v = v + u (commutative property)
4. u + (v + w) = (u + v) + w (associative property)
5. There exists an element of V, called the zero vector, denoted 0, such that u + 0 = u.
6. For every element u of V there exists an element called the negative of u, denoted u, such that u + (u) = 0.
• Scalar Multiplication Axioms
7. c(u + v) = cu + cv
8. (c + d)u = cu + du
9. c(du) = (cd)u
10. 1u = u
Definition of Vector Space (continued)
Ch04_29
W. Thus W is closed under scalar multiplication.
A Vector Space in R3
Let , for some a, bR.Wba )1,0,1( and )1,0,1( vu
Axiom 1:
u + v W. Thus W is closed under addition.
)1,0,1)(()1,0,1()1,0,1( baba vu
Axiom 3,4 and 7~10: trivial
}. | 1) 0, (1, {Let R aaW Prove that W is a vector space.
Proof
)1,0,1(cac uAxiom 2:
Axiom 5: Let 0 = (0, 0, 0) = 0(1,0,1), then 0 W and 0+u = u+0 = u for any u W.
Axiom 6: For any u = a(1,0,1) W. Let u = a(1,0,1), then u W and (u)+u = 0.
Ch04_30
Vector Spaces of Matrices (Mmn)
Let . and 22Mhg
fe
dc
ba
vu
Axiom 1:
u + v is a 2 2 matrix. Thus M22 is closed under addition.
► Question: Prove Axiom 2 and Axiom 7 .
hdgcfbea
hgfe
dcba
vu
Axiom 3 and 4:
From our previous discussions we know that 2 2 matrices are commutative and associative under addition (Theorem 2.2).
}.,,, | {Let 22 R
srqp
sr
qpM Prove that M22 is a vector space.
Proof
Ch04_31
{ | , , , 0} a vector space?p q
Is the set W p q r sr s
Axiom 5:
The 2 2 zero matrix is , since
00
000
u0u
dcba
dcba
0000
Axiom 6:
0uu
uu
0000
ddccbbaa
dcba
dcba
dcba
dcba
)(
since , then , If
In general: The set of m n matrices, Mmn, is a vector space.
Ch04_32
Vector Spaces of Functions
Axiom 1:
f + g is defined by (f + g)(x) = f(x) + g(x). f + g : R R
f + g F. Thus F is closed under addition.
Axiom 2:
cf is defined by (cf)(x) = c f(x).
cf : R R
cf F. Thus F is closed under scalar multiplication.
Prove that F = { f | f: R R } is a vector space.
Let f, g F, c R. For example: f: R R, f(x)=2x, g: R R, g(x)=x2+1.
Ch04_33
Vector Spaces of Functions (continued)Axiom 5:
Let 0 be the function such that 0(x) = 0 for every xR.
0 is called the zero function.
We get (f + 0)(x) = f(x) + 0(x) = f(x) + 0 = f(x) for every xR.
Thus f + 0 = f. (0 is the zero vector.)
Axiom 6:Let the function –f defined by (f )(x) = f (x).
Thus [f + (f )] = 0, f is the negative of f.)(
0
)]([)(
))(()())](([
x
xfxf
xfxfxff
0
Is the set F ={ f | f(x)=ax2+bx+c for some a,b,c R} a vector space?
Ch04_34
Theorem 4.4 (useful properties)
Let V be a vector space, v a vector in V, 0 the zero vector of V,
c a scalar, and 0 the zero scalar. Then
(a) 0v = 0
(b) c0 = 0
(c) (1)v = v(d) If cv = 0, then either c = 0 or v = 0.
Ch04_35
Homework
Exercise set 4.1, pages 206-207:5, 6, 13, 14.
Ch04_36
4.4 Subspaces
Figure 4.9
Ch04_37
Definition Let V be a vector space and U be a nonempty subset of V.
U is said to be a subspace of V if it is closed under addition
and under scalar multiplication.
Note:► In general, a subset of a vector space may or may not satisfy the closure axioms. ► However, any subset that is closed under both of these operations satisfies all the other vector space properties.
Ch04_38
Example 1Let U be the subset of R3 consisting of all vectors of the form (a, 0, 0) (with zeros as second and third components and aR ), i.e., U = {(a, 0, 0) R3 }.Show that U is a subspace of R3.
Solution
Let (a, 0, 0), (b, 0, 0) U, and let k R. We get
(a, 0, 0) + (b, 0, 0) = (a + b, 0, 0) Uk(a, 0, 0) = (k a, 0, 0) U
The sum and scalar product are in U. Thus U is a subspace of R3. #
Geometrically, U is the set of vectors that lie on the x-axis.
Ch04_39
Example 2Let V be the set of vectors of of R3 of the form (a, a2, b), namely V = {(a, a2, b) R3 }.Show that V is not a subspace of R3.
Solution
Let (a, a2, b), (c, c2, d) V. (a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)
(a + c, (a + c)2, b + d) ,
since a2 + c2 (a + c)2 .Thus (a, a2, b) + (c, c2, d) V. V is not closed under addition. V is not a subspace.
Ch04_40
Example 3Prove that the set W of 2 2 diagonal matrices is a subspace of the vector space M22 of 2 2 matrices.
Solution
(+) Let W.
qp
ba
00
and 0
0vu
We get
u + v W. W is closed under addition.
qbpa
qp
ba
00
00
00
vu
() Let c R. We get
cu W. W is closed under scalar multiplication. W is a subspace of M22.
cbca
ba
cc0
0
00
u
Ch04_41
The vector space of polynomials (Pn)
Example 5. Let Pn denoted the set of real polynomial functions of degree n. Prove that Pn is a vector space if addition and scalar multiplication are defined on polynomials in a pointwise manner.
Example 5. Let Pn denoted the set of real polynomial functions of degree n. Prove that Pn is a vector space if addition and scalar multiplication are defined on polynomials in a pointwise manner.
SolutionLet f and g Pn, where
11 1 0
11 1 0
( ) ... and
( ) ...
n nn n
n nn n
f x a x a x a x a
g x b x b x b x b
►(+)
(f + g)(x) is a polynomial of degree n. Thus f + g Pn. Then Pn is closed under addition.
)()(...)()(
]...[]...[
)()(
))((
00111
11
011
1011
1
baxbaxbaxba
bxbxbxbaxaxaxa
xgxf
xgf
nnn
nnn
nn
nn
nn
nn
Example 4.
Ch04_42
►() Let c R,
(cf )(x) is a polynomial of degree n. So cf Pn. Then Pn is closed under scalar multiplication.
In conclusion : By (+) and (), Pn is a subspace of the vector space F of functions. Therefore Pn is itself a vector space.
011
1
011
1
...
]...[
)]([))((
caxcaxcaxca
axaxaxac
xfcxcf
nn
nn
nn
nn
Ch04_43
Theorem 4.5 (Very important condition)
Let U be a subspace of a vector space V. U contains the zero vector of V.
Note. Let 0 be the zero vector of V. If 0 U U is not a subspace of V. If 0 U (+)() hold U is a subspace of V. (+)() failed U is not a subspace of V.Caution. This condition is necessary but not sufficient. (See, for instance, Example 2 above and Example 5 below)
Ch04_44
Example 5
Let W be the set of vectors of the form (a, a, a+2). Show that W is not a subspace of R3.
Solution
If (a, a, a+2) = (0, 0, 0), then a = 0 and a + 2 = 0 .
This system is inconsistent it has no solution.
Thus (0, 0, 0) W. (The necessary condition does not hold)
W is not a subspace of R3.
Ch04_45
Homework
Exercise set 1.3, page 32: 11, 12, 13.
Let F ={ f | f: R R } the vector space of functions on R.Which of the following are subspaces of F ?(a) W1={ f | f: R R, f(0)=0 }. (b) W2={ f | f: R R, f(0)=3 }. (c) W3={ f | f: R R, for some cR, f(x)=c for every x}.
Exercise
Ch04_46
4.5 Linear Combinations of Vectors
W={(a, a, b) | a,b R} R3
(a, a, b) = a (1,1,0) + b (0,0,1)
W is generated by (1,1,0) and (0,0,1). e.g., (2, 2, 3) = 2 (1,1,0) + 3 (0,0,1) (1, 1, 7) = 1 (1,1,0) + 7 (0,0,1).
Definition Let v1, v2, …, vm be vectors in a vector space V.
We say that v, a vector of V, is a linear combination of
v1, v2, …, vm , if there exist scalars c1, c2, …, cm such that
v can be write v = c1v1 + c2v2 + … + cmvm .
Ch04_47
The vector (5, 4, 2) is a linear combination of the vectors
(1, 2, 0), (3, 1, 4), and (1, 0, 3), since it can be written
(5, 4, 2) = (1, 2, 0) + 2(3, 1, 4) – 2(1, 0, 3)
Example 1
Ch04_48
Example 2
Determine whether or not the vector (1, 1, 5) is a linear combination of the vectors (1, 2, 3), (0, 1, 4), and (2, 3, 6).
Solution
Suppose 5) 1, 1,(6) 3, (2,4) 1, (0,3) 2, (1, 321 ccc
)5 ,1 ,1()643 ,32 ,2()5 ,1 ,1()6 ,3 ,2()4 , ,0()3 ,2 ,(
32132131
33322111
cccccccccccccccc
1,2,1
5 643
1 3 2
12
321
321
321
31
ccc
ccc
ccc
cc
Thus (1, 1, 5) is a linear combination of (1, 2, 3), (0, 1, 4), and (2, 3, 6), where 6). 3, (2,14) 1, (0,23) 2, (1,5) 1, 1,(
Ch04_49
Example 3
Express the vector (4, 5, 5) as a linear combination of the vectors (1, 2, 3), (1, 1, 4), and (3, 3, 2).
Solution
Suppose 5) 5, (4,2) 3, (3,4) 1,1(3) 2, (1, 321 c, cc
)5 ,5 ,4()243 ,32 ,3()5 ,5 ,4()2 ,3 ,3()4 , ,()3 ,2 ,(
321321321
333222111
cccccccccccccccccc
rcrcrc
ccc
ccc
ccc
321
321
321
321
,1,32
5 243
53 2
43
Thus (4, 5, 5) can be expressed in many ways as a linear combination of (1, 2, 3), (1, 1, 4), and (3, 3, 2):
2) 3, ,3(4) 1, ,1()1(3) 2, (1,)32(5) 5, ,4( rrr
Ch04_50
Example 4
Show that the vector (3, 4, 6) cannot be expressed as a linear combination of the vectors (1, 2, 3), (1, 1, 2), and (1, 4, 5).
Solution
Suppose
)6 ,4 ,3(5) 4, (1,2) ,11(3) 2, (1, 321 c, cc
6 523
44 2
3
321
321
321
ccc
ccc
ccc
This system has no solution. Thus (3, 4, 6) is not a linear combination of the vectors (1, 2, 3), (1, 1, 2), and (1, 4, 5).
Ch04_51
Example 5
Determine whether the matrix is a linear combination
of the matrices in the vector space
M22 of 2 2 matrices.
1871
0210
and ,2032
,1201
SolutionSuppose
1871
0210
2032
1201
321 ccc
18
71
222
32
2121
3221
cccc
ccccThen
Ch04_52
12
822
73
12
21
31
32
21
cc
cc
cc
cc
This system has the unique solution c1 = 3, c2 = 2, c3 = 1. Therefore
0210
2032
21201
318
71
Ch04_53
Example 6Determine whether the function is a linear combination of the functions and
710)( 2 xxxf
Solution
Suppose .21 fhcgc
710)42()13( 222
21 xxxxcxxc
Then
13)( 2 xxxg
.42)( 2 xxxh
7104)3()2( 22121
221 xxccxccxcc
74
103
12
21
21
21
cc
cc
cc
1 ,3 21 cc .3 hgf
Ch04_54
Definition The vectors v1, v2, …, vm are said to span a vector space if every vector in the space can be expressed as a linear combination of these vectors.
In this case {v1, v2, …, vm} is called a spanning set.
Spanning Sets
Ch04_55
Show that the vectors (1, 2, 0), (0, 1, 1), and (1, 1, 2) span R3.
SolutionLet (x, y, z) be an arbitrary element of R3. Suppose )2 ,1 ,1()1 ,1 ,0()0 ,2 ,1() , ,( 321 ccczyx
)2 ,2 ,() , ,( 3232131 ccccccczyx
Example 7
zcc
yccc
xcc
32
321
31
2
2
zyxc
zyxc
zyxc
2
24
3
3
2
1
The vectors (1, 2, 0), (0, 1, 1), and (1, 1, 2) span R3.
)2 ,1 ,1)(2()1 ,1 ,0)(24()0,2,1)(3() , ,( zyxzyxzyxzyx
Ch04_56
Example 8Show that the following matrices span the vector space M22 of 2 2 matrices.
1000
0100
0010
0001
Solution
dcba
Let M22 (an arbitrary element).
We can express this matrix as follows:
proving the result.
1000
0100
0010
0001
dcbadcba
Ch04_57
Theorem 4.6
Let v1, …, vm be vectors in a vector space V. Let U be the set consisting of all linear combinations of v1, …, vm . Then U is a subspace of V spanned the vectors v1, …, vm . U is said to be the vector space generated by v1, …, vm .Proof
(+) Let u1 = a1v1 + … + amvm and u2 = b1v1 + … + bmvm U.
Then u1 + u2 = (a1v1 + … + amvm) + (b1v1 + … + bmvm) = (a1 + b1) v1 + … + (am + bm) vm
u1 + u2 is a linear combination of v1, …, vm .
u1 + u2 U.
U is closed under vector addition.
Ch04_58
()Let c R. Then
cu1 = c(a1v1 + … + amvm)
= ca1v1 + … + camvm
cu1 is a linear combination of v1, …, vm . cu1 U.
U is closed under scalar multiplication.
Thus U is a subspace of V.
By the definition of U, every vector in U can be written as a linear combination of v1, …, vm .
Thus v1, …, vm span U.
Ch04_59
Example 9
Consider the vector space R3.
The vectors (1, 5, 3) and (2, 3, 4) are in R3.
Let U be the subset of R3 consisting of all vectors of the form
c1(1, 5, 3) + c2(2, 3, 4)
Then U is a subspace of R3 spanned by (1, 5, 3) and (2, 3, 4).
The following are examples of some of the vectors in U, obtained by given c1 and c2 various values.
)81 ,1 ,4(vector ;3 ,2)0 ,0 ,0(vector ;0 ,0
)4 ,3 ,2(vector ;1 ,0)3 ,5 ,1(vector ;0 ,1
21
21
21
21
cccccccc
We can visualize U. U is made up of all vectors in the plane defined by the vectors (1, 5, 3) and (2, 3, 4).
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Figure 4.10
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We can generalize this result. Let v1 and v2 be vectors in the space R3.
The subspace U generated by v1 and v2 is the set of all vectors of the form c1v1 + c2v2.
If v1 and v2 are not colinear, then U is the plane defined by v1 and v2 .
Figure 4.11
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Figure 4.12
If v1 and v2 are vectors in R3 that are not colinear, then we can visualize U as a plane in three dimensions.
k1v1 and k2v2 will be vectors on the same lines as v1 and v2.
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Example 10Let U be the subspace of R3 generated by the vectors (1, 2, 0) and (3, 1, 2). Let V be the subspace of R3 generated by the vectors (1, 5, 2) and (4, 1, 2). Show that U = V.
(U V) Let u be a vector in U. Let us show that u is in V. Since u is in U, there exist scalars a and b such that
Let us see if we can write u as a linear combination of (1, 5, 2) and (4, 1, 2).
Such p and q would have to satisfy
Solution
)2 ,2 ,3()2 ,1 ,3()0 ,2 ,1( bbababa u
)22 ,5 ,4()2- ,1 ,4()2 ,5 ,1( qpqpqpqp u
bqpbaqpbaqp
22225
34
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This system of equations has unique solutionThus u can be written
Therefore, u is a vector in V.
.32
,3
baq
bap
)2 ,1 ,4(32
)2 ,5 ,1(3
babau
(V U) Let v be a vector in V. Let us show that v is in U. Since v is in V, there exist scalars c and d such that
It can be shown that)2 14()2 ,5 ,1( ,, dcv
)2 ,1 ,3)(()0 ,2 ,1)(2( dcdcv
Therefore, v is a vector in U and hence U=V.
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Figure 4.13
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Example 11Let U be the vector space generated by the functions f(x) = x + 1 and g(x) = 2x2 – 2x + 3. Show that the function h(x) = 6x2–10x+5 lies in U.Solutionh will be in the space generated by f and g if there exist scalars a and b such that
a(x + 1) + b(2x2 – 2x + 3) = 6x2 – 10x + 5 This given
2bx2 + (a – 2b)x + a + 3b = 6x2 – 10x + 5
2b = 6 a – 2b = – 10
a + 3b = 5This system has the unique solution a = – 4, b = 3. Thus – 4(x + 1) + 3(2x2 – 2x + 3) = 6x2 – 10x + 5
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Homework
Exercise set 4.2 pages 214-215:2, 4, 6, 7, 15, 17.
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4.6 Linear Dependence and Independence
Definition(a) The set of vectors { v1, …, vm } in a vector space V is said to
be linearly dependent if there exist scalars c1, …, cm, not all zero, such that c1v1 + … + cmvm = 0
(b) The set of vectors { v1, …, vm } is linearly independent if c1v1 + … + cmvm = 0 can only be satisfied when
c1 = 0, …, cm = 0.
The concepts of dependence and independence of vectors are useful tools in constructing “efficient” spanning sets for vectorspaces – sets in which there are no redundant vectors.
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Example 1
Show that the set {(1, 2, 3), (2, 1, 1), (8, 6, 10)} is linearly dependent in R3.
SolutionSuppose
0 )10 ,6 ,8()1 ,1 ,2()3 ,2 ,1( 321 ccc
00
)103 ,62 ,82()10 ,6 ,8() , ,2()3 ,2 ,(
321321321
333222111
cccccccccccccccccc
Thus The set of vectors is linearly dependent.
0 )10 ,6 ,8()1 ,1 ,2(2)3 ,2 ,1(4
c1 = 4c2 = 2c3 = 1
0103
062
082
321
321
321
ccc
ccc
ccc
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Example 2
Show that the set {(3, 2, 2), (3, 1, 4), (1, 0, 5)} is linearly independent in R3.
Suppose
0 )5 ,0 ,1()4 ,1 ,3()2 ,2 ,3( 321 ccc
00
)542 ,2 ,33()5 ,0 ,()4 , ,3()2 ,2 ,3(
32121321
33222111
cccccccccccccccc
Solution
This system has the unique solution c1 = 0, c2 = 0, and c3 = 0. Thus the set is linearly independent.
0542
02
033
321
21
321
ccc
cc
ccc
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Example 3Consider the functions f(x) = x2 + 1, g(x) = 3x – 1, h(x) = – 4x + 1 of the vector space P2 of polynomials of degree 2. Show that the set of functions { f, g, h } is linearly independent.Solution
Supposec1f + c2g + c3h = 0
Since for any real number x,
Consider three convenient values of x. We get
0 )14()13()1( 322
1 xcxcxc
0542 :10322 :1
0 :0
321
321
321
cccxcccx
cccx
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It can be shown that this system of three equations has the unique solution
c1 = 0, c2 = 0, c3 = 0
Thus c1f + c2g + c3h = 0 implies that c1 = 0, c2 = 0, c3 = 0. The set { f, g, h } is linearly independent.
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Theorem 4.7
A set consisting of two or more vectors in a vector space is linearly dependent if and only if it is possible to express one of the vectors as a linearly combination of the other vectors.
Example 4The set of vectors {v1=(1, 2, 1) , v2 =(-1, -1, 0) , v3 = (0, 1,1)}
is linearly dependent, since v3 = v1 + v2 .Thus, v3 is a linear combination of v1 and v2.
m
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Linear Dependence of {v1, v2}
{v1, v2} linearly dependent;vectors lie on a line
{v1, v2} linearly independent;vectors do not lie on a line
Figure 4.14 Linear dependence and independence of {v1, v2} in R3.
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Linear Dependence of {v1, v2, v3}
{v1, v2, v3} linearly dependent;vectors lie in a plane
{v1, v2, v3} linearly independent;vectors do not lie in a plane
Figure 4.15 Linear dependence and independence of {v1, v2, v3} in R3.
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Theorem 4.8
Let V be a vector space. Any set of vectors in V that contains the zero is linearly dependent.
Proof
Consider the set { 0, v2, …, vm }, which contains the zero vectors. Let us examine the identity
0vv0 mmccc 221
We see that the identity is true for c1 = 1, c2 = 0, …, cm = 0 (not all zero). Thus the set of vectors is linearly dependent, proving the theorem.
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Theorem 4.9Let the set {v1, …, vm} be linearly dependent in a vector space V. Any set of vectors in V that contains these vectors will also be linearly dependent.
Example 5The set of vectors
{v1=(1, 2, 1) , v2 =(-1, -1, 0) , v3 = (0, 1,1) , v4 =( 1, 1, 1) } is linearly dependent, since it contains the vectors v1 , v2 , v3 which are linearly dependent.
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Example 6
Let the set {v1, v2} be linearly independent. Prove that {v1 + v2, v1 – v2} is also linearly independent.
Solution
Suppose(1)
We get0)()( 2121 vvvv ba
0)()(0
21
2121
vvvvvv
bababbaa
Since {v1, v2} is linearly independent
Thus system has the unique solution a = 0, b = 0. Returning to identity (1) we get that {v1 + v2, v1 – v2} is linearly independent.
00
baba
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Homework
Exercise set 4.3 pages 219-220:1, 3, 8, 9, 12, 16.
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4.7 Bases and Dimension
DefinitionA finite set of vectors {v1, …, vm} is called a basis for a vector space V if the set spans V and is linearly independent.
Standard BasisThe set of n vectors
{(1, 0, …, 0), (0, 1, …, 0), …, (0, …, 1)}
is a basis for Rn. This basis is called the standard basis for Rn.
How to prove it?
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Example 1
Show that the set {(1, 0, 1), (1, 1, 1), (1, 2, 4)} is a basis for R3.
Solution
(span)Let (x1, x2, x3) be an arbitrary element of R3. Suppose
)4 ,2 ,1()1 ,1 ,1()1 ,0 ,1() , ,( 321321 aaaxxx
Thus the set spans the space.
3321
232
1321
4
2
xaaa
xaa
xaaa
3213
3212
3211
2
252
32
xxxa
xxxa
xxxa
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04020
321
32
321
bbbbbbbb
(linearly independent) Consider the identity
The identity leads to the system of equations)0 ,0 ,0()4 ,2 ,1()1 ,1 ,1()1 ,0 ,1( 321 bbb
b1 = 0, b2 = 0, and b3 = 0 is the unique solution. Thus the set is linearly independent. {(1, 0, 1), (1, 1, 1), (1, 2, 4)} spans R3 and is linearly independent. It forms a basis for R3.
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Example 2
Show that { f, g, h }, where f(x) = x2 + 1, g(x) = 3x – 1, and h(x) = –4x + 1 is a basis for P2.
Solution(linearly independent) see Example 3 of the previous section. (span). Let p be an arbitrary function in P2. p is thus a polynomial of the form
dcxbxxp 2)(
Suppose )()()()( 321 xhaxgaxfaxp
for some scalars a1, a2, a3.
This gives
)()43(
)14()13()1(
321322
1
322
12
aaaxaaxa
xaxaxadcxbx
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Thus the polynomial p can be expressed
The functions f, g, and h span P2.
They form a basis for P2.
)()()()( 321 xhaxgaxfaxp
daaacaaba
321
32
1
43cdba
cdba
ba
33
44
3
2
1
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Theorem 4.10 Let {v1, …, vn } be a basis for a vector space V. If {w1, …, wm} is a set of more than n vectors in V, then this set is linearly dependent.Proof
Suppose(1)
We will show that values of c1, …, cm are not all zero.
011 mmcc ww
The set {v1, …, vn} is a basis for V. Thus each of the vectors w1, …, wm can be expressed as a linear combination of v1, …, vn. Let
nmnmmm
nn
aaa
aaa
vvvw
vvvw
2211
12121111
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Substituting for w1, …, wm into Equation (1) we get
0vvvvvv )()( 221112121111 nmnmmmnn aaacaaac
0vv nmnmnnmm acacacacacac )()( 221111212111
Since v1, …, vn are linear independent,
0
0
2211
1221111
mmnnn
mm
cacaca
cacaca
Since m > n, there are many solutions in this system.
Rearranging, we get
Thus the set {w1, …, wm} is linearly dependent.
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Theorem 4.11Any two bases for a vector space V consist of the same number of vectors.
Proof
Let {v1, …, vn} and {w1, …, wm} be two bases for V.
Thus n = m.
By Theorem 4.11, m n and n m
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DefinitionIf a vector space V has a basis consisting of n vectors, then the dimension of V is said to be n. We write dim(V) for the dimension of V.
• V is finite dimensional if such a finite basis exists.• V is infinite dimensional otherwise.
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Example 3
Consider the set {{1, 2, 3), (-2, 4, 1)} of vectors in R3. These vectors generate a subspace V of R3 consisting of all vectors of the form
The vectors (1, 2, 3) and (-2, 4, 1) span this subspace.
)1 ,4 ,2()3 ,2 ,1( 21 ccv
Furthermore, since the second vector is not a scalar multiple of the first vector, the vectors are linearly independent.Therefore {{1, 2, 3), (-2, 4, 1)} is a basis for V. Thus dim(V) = 2. We know that V is, in fact, a plane through the origin.
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Theorem 4.12
(a) The origin is a subspace of R3. The dimension of this subspace is defined to be zero.
(b) The one-dimensional subspaces of R3 are lines through the origin.
(c) The two-dimensional subspaces of R3 are planes through the origin.
Figure 4.16 One and two-dimensional subspaces of R3
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Proof
(a) Let V be the set {(0, 0, 0)}, consisting of a single elements, the zero vector of R3. Let c be the arbitrary scalar. Since
(0, 0, 0) + (0, 0, 0) = (0, 0, 0) and c(0, 0, 0) = (0, 0, 0)
V is closed under addition and scalar multiplication. It is thus a subspace of R3. The dimension of this subspaces is defined to be zero.
(b) Let v be a basis for a one-dimensional subspace V of R3. Every vector in V is thus of the form cv, for some scalar c. We know that these vectors form a line through the origin.
(c) Let {v1, v2}be a basis for a two-dimensional subspace V of R3. Every vector in V is of the form c1v1 + c2v2. V is thus a plane through the origin.
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Theorem 4.13
Let {v1, …, vn} be a basis for a vector space V. Then each vector in V can be expressed uniquely as a linear combination of these vectors.
ProofLet v be a vector in V. Since {v1, …, vn} is a basis, we can express v as a linear combination of these vectors. Suppose we can write
nnnn bbaa vvvvvv 1111 and Then
givingnnnn bbaa vvvv 1111
0vv nnn baba )()( 111 Since {v1, …, vn} is a basis, the vectors v1, …, vn are linearly independent. Thus (a1 – b1) = 0, …, (an – bn) = 0, implying that a1 = b1, …, an = bn. There is thus only one way of expressing v as a linear combination of the basis.
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Theorem 4.14
Let V be a vector space of dimension n.
(a) If S = {v1, …, vn} is a set of n linearly independent vectors in V, then S is a basis for V.
(b) If S = {v1, …, vn} is a set of n vectors V that spans V, then S is a basis for V.
Let V be a vector space, S = {v1, …, vn} is a set of vectors in V.
(a) dim(V) = |S|.
(b) S is a linearly independent set.
(c) S spans V.S is a basis of V.
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Example 4Prove that the set B={(1, 3, -1), (2, 1, 0), (4, 2, 1)} is a basis for R3.
SolutionSince dim(R3)=|B|=3. It suffices to show that this set is linearly independent or it spans R3.Let us check for linear independence. Suppose
)0 ,0 ,0()1 ,2 ,4()0 ,1 ,2()1 ,3 ,1( 321 ccc
This identity leads to the system of equations
This system has the unique solution c1 = 0, c2 = 0, c3 = 0. Thus the vectors are linearly independent.The set {(1, 3, -1), (2, 1, 0), (4, 2, 1)} is therefore a basis for R3.
0023042
31
321
321
cccccccc
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Theorem 4.15
Let V be a vector space of dimension n. Let {v1, …, vm} be a set of m linearly independent vectors in V, where m < n.
Then there exist vectors vm+1, …, vn such that
{v1, …, vm, vm+1, …, vn } is a basis of V.
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Example 5State (with a brief explanation) whether the following statements
are true or false.
(a) The vectors (1, 2), (1, 3), (5, 2) are linearly dependent in R2.
(b) The vectors (1, 0, 0), (0, 2, 0), (1, 2, 0) span R3.
(c) {(1, 0, 2), (0, 1, -3)} is a basis for the subspace of R3 consisting of vectors of the form (a, b, 2a 3b).
(d) Any set of two vectors can be used to generate a two-dimensional subspace of R3.
Solution
(a) True: The dimension of R2 is two. Thus any three vectors are linearly dependent.
(b) False: The three vectors are linearly dependent. Thus they cannot span a three-dimensional space.
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(c) True: The vectors span the subspace since
(a, b, 2a, -3b) = a(1, 0, 2) + b(0, 1, -3)
The vectors are also linearly independent since they are not colinear.
(d) False: The two vectors must be linearly independent.
Homework
Exercise set 4.4 pages 226-227:5, 7, 11, 15, 18, 20, 21, 31.
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