chapter 4 general vector spaces 大葉大學 資訊工程系 黃鈴玲 linear algebra
TRANSCRIPT
Chapter 4General Vector Spaces
大葉大學 資訊工程系黃鈴玲
Linear Algebra
Ch04_2
4.1 General Vector Spaces and Subspaces
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. (u, v, and w are arbitrary elements of V, and c and d are scalars.)
Closure Axioms (最重要！ )
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. We draw up a set of axioms ( 公理 ) based on the properties of Rn. Any set that satisfies these axioms will have similar algebraic properties to the vector space Rn.
Ch04_3
補充範例(1) V={ …, 3, 1, 1, 3, 5, 7, …} ( 所有奇數構成的集合 ) V is not closed under addition because 1+3=4 V.
(2) Z={ …, 2, 1, 0, 1, 2, 3, 4, …} ( 所有整數構成的集合 ) 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.
隨堂作業： 14
Ch04_4
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_5
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.補充：
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. 隨堂作業： 5
Ch04_6
Vector Spaces of Matrices
Let . and 22Mhg
fe
dc
ba
vu
Axiom 1:
u + v is a 2 2 matrix. Thus M22 is closed under addition.
hdgcfbea
hgfe
dcba
vu
Axiom 3 and 4:
From our previous discussions we know that 2 2 matrices are communicative and associative under addition (Theorem 2.2).
}.,,, | {Let 22 R
srqp
sr
qpM Prove that M22 is a vector space.
Proof
Ch04_7
}.0,,, | {
srqp
sr
qpW
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
推廣： The set of m n matrices, Mmn, is a vector space.
隨堂作業： 7
Ch04_8
Vector Spaces of Functions
Axiom 1:
f + g is defined by (f + g)(x) = f(x) + g(x). f + g : R R
f + g V. Thus V is closed under addition.
Axiom 2:
cf is defined by (cf)(x) = c f(x).
cf : R R
cf V. Thus V is closed under scalar multiplication.
Prove that V={ f | f: R R } is a vector space.
Let f, g V, c R. For example: f: R R, f(x)=2x, g: R R, g(x)=x2+1.
( 跳過 )
Ch04_9
Vector Spaces of FunctionsAxiom 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
V={ f | f(x)=ax2+bx+c for some a,b,c R}隨堂作業： 13
( 跳過 )
Ch04_10
The Complex Vector Space Cn
) ..., ,() ..., ,(
) ..., ,() ..., ,() ..., ,(
11
1111
nn
nnnn
cucuuuc
vuvuvvuu
:follows as on defined be )scalar complex a(by
tion multiplicascalar andaddition of operationsLet
. denoted is sequencessuch all ofset The
numbers.complex of sequence a be ) ..., ,(Let 1
n
n
n
c
nuu
C
C
It can be shown that Cn with these two operations is acomplex vector space.
( 跳過 )
Ch04_11
Theorem 4.1Let 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.
(a) 0v + 0v = (0 + 0)v = 0v (0v + 0v) + (0v) = 0v + (0v) 0v + [0v + (0v)] = 0, 0v + 0 = 0, 0v = 0(c) (1)v + v = (1)v + 1v = [(1) + 1]v = 0v = 0
Proof
Ch04_12
Definition Let V be a vector space and U be a nonempty subset of V. If U is a vector space under the operations of addition and scalar multiplication of V it is called a subspace of V.
In general, a subset of a vector space may or may not satisfy the closure axioms. However, any subset that is closed under bothof these operations satisfies all the other vector space properties.
Subspaces
Note. U 只要有加法與純量乘法的封閉性，其他 axiom 都會滿足。
Ch04_13
Example 1Let W be the subset of R3 consisting of all vectors of the form (a, a, b). Show that W is a subspace of R3.
Solution
Let (a, a, b), (c, c, d) W, and let k R. We get
(a, a, b) + (c, c, d) = (a+c, a+c, b+d) Wk(a, a, b) = (ka, ka, kb) W
Thus W is a subspace of R3.
Ch04_14
Example 1’Let W be the set of vectors of the form (a, a2, b). Show that W is not a subspace of R3.
Solution
Let (a, a2, b), (c, c2, d) W. (a, a2, b) + (c, c2, d) = (a+ c, a2 + c2, b + d)
(a + c, (a + c)2, b + d)Thus (a, a2, b) + (c, c2, d) W. W is not closed under addition. W is not a subspace.
隨堂作業： 18(a), 20(b)
Ch04_15
Example 2Prove that the set U of 2 2 diagonal matrices is a subspace of the vector space M22 of 2 2 matrices.
Solution
(+) Let U.
qp
ba
00
and 0
0vu
We get
u + v U. U is closed under addition.
qbpa
qp
ba
00
00
00
vu
() Let c R. We get
cu U. U is closed under scalar multiplication. U is a subspace of M22.
cbca
ba
cc0
0
00
u
隨堂作業： 27(a)
Ch04_16
Example 3Let 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
011
1
011
1
...)(
and ...)(
bxbxbxbxg
axaxaxaxfn
nn
n
nn
nn
(+)
(f + g)(x) is a polynomial of degree n. Thus f + g Pn. Pn is closed under addition.
)()(...)()(
]...[]...[
)()(
))((
00111
11
011
1011
1
baxbaxbaxba
bxbxbxbaxaxaxa
xgxf
xgf
nnn
nnn
nn
nn
nn
nn
( 跳過 )
Ch04_17
() Let c R,
(cf)(x) is a polynomial of degree n. Thus cf Pn. Pn is closed under scalar multiplication.
By (+) and (), Pn is a subspace of the vector space V of functions. Therefore Pn is a vector space.
011
1
011
1
...
]...[
)]([))((
caxcaxcaxca
axaxaxac
xfcxcf
nn
nn
nn
nn
( 跳過 )
Ch04_18
Theorem 4.2Let U be a subspace of a vector space V. U contains the zero vector of V.
Proof
Let u be an arbitrary vector in U and 0 be the zero vector of V. Let 0 be the zero scalar.
By Theorem 4.5(a) we know that 0u = 0.
Since U is closed under scalar multiplication, this means that 0 is in U.
Note. Let 0 be the zero vector of V, U is a subset of V. If 0 U U is not a subspace of V. If 0 U check (+)() to determine if U is a subspace of V.
Ch04_19
Example 4Let 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.
(0, 0, 0) W.
W is not a subspace of R3.
隨堂作業： 24(a,b), 27(b)
Ch04_20
Homework
Exercise 4.1:5, 7, 14, 18, 19, 20, 24, 27
Ch04_21
4.2 Linear CombinationsW={(a, a, b) | a,b R} R3
(a, a, b) = a (1,1,0) + b (0,0,1)
W 中的任何向量都可以用 (1,1,0) 及 (0,0,1) 來表示e.g., (2, 2, 3) = 2 (1,1,0) + 3 (0,0,1) (1, 1, 7) = 1 (1,1,0) + 7 (0,0,1)
W 中的任何向量都是 (1,1,0) 及 (0,0,1) 的 linear combination.
Definition Let v1, v2, …, vm be vectors in a vector space V. The vector v in V is a linear combination ( 線性組合 ) of v1, v2, …, vm if there exist scalars c1, c2, …, cm such that v can be written
v = c1v1 + c2v2 + … + cmvm
Ch04_22
The vector (7, 3, 2) is a linear combination of the vector (1, 3, 0) and (2, 3, 1) since
(7, 3, 2) = 3(1, 3, 0) + 2(2, 3, 1)
The vector (3, 4, 2) is not a linear combination of (1, 1, 0) and (2, 3, 0) because there are no values of c1 and c2 for which
(3, 4, 2) = c1(1, 1, 0) + c2(2, 3, 0)
is true.
Example
Ch04_23
Example 1Determine whether or not the vector (, 0, 5) is a linear combination of the vectors (1, 2, 3), (0, 1, 4), and (2, 1, 1).
Solution
Suppose c1(1, 2, 3)+c2(0, 1, 4)+c3(2, 1, 1)=(8, 0, 5).
5 430 282
321
321
31
cccccccc
Thus (8, 0, 5) is a linear combination of (1, 2, 3), (0, 1, 4), and (2, 1, 1).
隨堂作業： 2(a)
3,1,2 321 ccc
Ch04_24
決定一個向量是否是某些向量的 linear combination 求解聯立方程式 唯一解表示 linear combination 的係數唯一 無限多解表示 linear combination 的係數不唯一 無解表示不是 linear combination
Ch04_25
Example 2Determine whether the vector (4, 5, 5) is 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
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):
6) 3, ,2(4) 1, ,1()1(3) 2, (1,)32(5) 5, ,4( rrr
Ch04_26
Example 2’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 (1, 2, 3),(1, 1, 2), and (1, 4, 5).
隨堂作業： 2(c)
Ch04_27
Definition Let v1, v2, …, vm be vectors in a vector space V. These vectors span V if every vector in V can be expressed as a linear combination of them.
{v1, v2, …, vm} is called a spanning set of V.
Spanning a Vector Space
Ch04_28
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 3
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
隨堂作業： 4(a)
Ch04_29
Theorem 4.3Let 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. It is denoted Span{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_30
()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_31
Example 4Consider the vectors (1, 5, 3) and (2, 3, 4) in R3. Let U =Span{(1, 5, 3), (2, 3, 4)}. U will be a subspace of R3 consisting of all vectors of the form
c1(1, 5, 3) + c2(2, 3, 4).
The following are examples of some of the vectors in U, which are obtained by given c1 and c2 various values.
)81 ,1 ,4(vector ;3 ,2
)7 ,2 ,1(vector ;1 ,1
21
21
cc
cc
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).
Ch04_32
Figure 4.1
Ch04_33
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 4.2
Ch04_34
Example 5Let v1 and v2 span a subspace U of a vector V. Let k1 and k2 be nonzero scalars. Show that k1v1 and k2v2 also span U.
Solution
Choose any vector v U. Since v1 and v2 span U, There exist a, b R such that v = av1 + bv2
We can write
k1v1 and k2v2 span U.
)()( 222
111
vvv kkb
kka
隨堂作業： 20
Ch04_35
Example 6Determine 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_36
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_37
Example 7Show that the function h(x) =4x2+3x7 lies in the space Span{f, g} generated by f(x) = 2x25 and g(x) = x+1.
Solution
Suppose .21 hgcfc 734)1()52( 2
22
1 xxxcxcThen
73452 2212
21 xxccxcxc
75
3
42
21
2
1
cc
c
c3 ,2 21 cc .32 gfh
( 跳過 )
Therefore the function h(x) lies in Span{f, g}.
Ch04_38
Homework
Exercise 4.2:2, 4, 6, 18, 20
Ch04_39
4.3 Linear Dependence and Independence
Definition(a) The set of vectors { v1, …, vm } in a vector space V is said to
be linearly dependent ( 線性相依 ) if there exist scalars c1, …, cm, not all zero, such that c1v1 + … + cmvm = 0
(b) The set of vectors { v1, …, vm } is linearly independent ( 線性獨立 ) 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.
Ch04_40
Example 1Determine whether the set{(1, 2, 0), (0, 1, 1), (1, 1, 2)} is linearly dependent in R3.
Solution
Suppose 0 )2 ,1 ,1()1 ,1 ,0()0 ,2 ,1( 321 ccc
Thus the set of vectors is linearly independent.
c1 = 0c2 = 0 is the unique solution.c3 =
02
02
0
32
321
31
cc
ccc
cc
Note. 不需真的求解，只需判斷是唯一解或無限多解，故當係數矩陣是方陣時，求算係數矩陣的行列式即可。此題行列式 0 唯一解 線性獨立 若行列式 =0 無限多解 線性相依
隨堂作業： 1(c,e)
Ch04_41
Example 2(a) Show that the set {x2+1, 3x–1, –4x+1} is linearly independent in P2.(b) Show that the set {x+1, x–1, –x+5} is linearly dependent in P1.Solution
(a) Suppose c1(x2 + 1) + c2(3x – 1) + c3(– 4x + 1) = 0 c1x2 +(3c2 – 4c3)x + c1 – c2 + c3 = 0 c1 = 0, c2 = 0, c3 = 0 is the unique solution linearly independent
( 跳過 )
(b) Suppose c1(x+1) + c2(x –1) + c3(–x+5) = 0 (c1+ c2 – c3)x + c1 – c2 +5c3 = 0 many solutions linearly dependent
Ch04_42
Theorem 4.4
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.
Proof() Let the set { v1, v2, …, vm } be linearly dependent. Therefore, there exist scalars c1, c2, …, cm, not all zero, such that
c1v1 + c2v2 + … + cmvm = 0Assume that c1 0. The above identity can be rewritten
Thus, v1 is a linear combination of v2, …, vm.
mm
c
c
cc
vvv
1
21
21
Ch04_43
() Conversely, assume that v1 is a linear combination of v2, …, vm. Therefore, there exist scalars d2, …, dm, such that
v1 = d2v2 + … + dmvm
Rewrite this equation as
1v1 + ( d2)v2 + … + (dm)vm = 0
Thus the set {v1, v2, …, vm} is linearly dependent, completing the proof.
Ch04_44
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 4.3 Linear dependence and independence of {v1, v2} in R3.
Ch04_45
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 4.4 Linear dependence and independence of {v1, v2, v3} in R3.
Ch04_46
Theorem 4.5Let 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.
Ch04_47
Theorem 4.6Let 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.
Proof
Since the set {v1, …, vm} is linearly dependent, there exist scalars c1, …, cm, not all zero, such that
0vv mmcc 11
Consider the set of vectors {v1, …, vm, vm+1, …, vn}, which contains the given vectors. There are scalars, not all zero, namely c1, …, cm, 0, …, 0 such that
Thus the set {v1, …, vm, vm+1, …, vn} is linearly dependent.
0vvvv nmmmcc 00 111
Ch04_48
Example 3Let the set {v1, v2} be linearly independent. Prove that {v1 + v2, v1 – v2} is also linearly independent.
Solution
Suppose(1)
We get
0vvvv )()( 2121 ba
0vv
0vvvv
21
2121
)()( baba
bbaa
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 隨堂作業： 12
Ch04_49
Homework
Exercise 4.3:1, 3, 8, 12
Ch04_50
4.4 Properties of Bases
Let the vectors v1, …, vn span a vector space V. Each vector in V can be expressed uniquely as a linear combination of these vectors if and only if the vectors are linearly independent.
Theorem 4.7
Proof
(a) () Assume that v1, …, vn are linearly independent. Let v V. Since v1, …, vn span V, we can express v as a linear combination of these vectors. Suppose we can write
Since v1, …, vn are linearly independent, a1 – b1 = 0, …, an – bn = 0, implying that a1 = b1, …, an = bn.
nnnn bbaa vvvv 1111
0vv nnn baba )()( 111
nnnn bbaa vvvvvv 1111 and
unique 得證
Ch04_51
(b) () Let v V. Assume that v can be written in only one way as a linear combination of v1, …, vn. Note that 0v1+ …+ 0vn= 0. If c1v1+ …+ cnvn= 0, it implies that c1 = 0, c2 = 0, …, cn = 0.
v1, …, vn are linearly independent.
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.
Ch04_52
Theorem 4.8 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.
ProofSuppose
(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
Ch04_53
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.
隨堂作業： 6(b)
Ch04_54
Theorem 4.9Any 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 4.8, m n and n m
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.
Ch04_55
Example 1
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.
Ch04_56
Theorem 4.10(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 4.5 One and two-dimensional subspaces of R3
Ch04_57
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.
隨堂作業： 16(a)
Ch04_58
Theorem 4.11Let 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.若以下三點有兩點成立，則另一點也成立。
(a) dim(V) = |S|.
(b) S is a linearly independent set.
(c) S spans V.
整理 :
S is a basis of V.
Ch04_59
Example 2Prove 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
隨堂作業： 5(a), 20(c), 21
Ch04_60
Theorem 4.12Let 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.
Ch04_61
Example 3State (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.
Ch04_62
(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.
Ch04_63
Homework
Exercise 4.4:5, 6, 7, 16, 20, 21, 23, 25
Ch04_64
4.5 Rank
DefinitionLet A be an m n matrix. The rows of A may be viewed as row vectors r1, …, rm, and the columns as column vectors c1, …, cn. Each row vector will have n components, and each column vector will have m components, The row vectors will span a subspace of Rn called the row space of A, and the column vectors will span a subspace of Rm called the column space of A.
Rank enables one to relate matrices to vectors, and vice versa.
Ch04_65
Example 1
Consider the matrix
(1) The row vectors of A are
014561432121
)0 ,1 ,4 ,5(),6 ,1 ,4 ,3( ),2 ,1 ,2 ,1( 321 rrr
(2) The column vectors of A are
These vectors span a subspace of R3 called the column space of A.
062
111
442
531
4321 cccc
These vectors span a subspace of R4 called the row space of A.
Ch04_66
Theorem 4.13The row space and the column space of a matrix A have the same dimension.
Proof
Let u1, …, um be the row vectors of A. The ith vector is
Let the dimension of the row space be s. Let the vectors v1, …, vs form a basis for the row space. Let the jth vector of this set be
) ..., , ,( 21 iniii aaau
) ..., , ,( 21 jnjjj bbbv
Each of the row vectors of A is a linear combination of v1, …, vs. Let
smsmmm
ss
ccc
ccc
vvvu
vvvu 1
2211
1212111
Ch04_67
Equating the ith components of the vectors on the left and right, we get
This may be writtensimsimimmi
sisiii
bcbcbca
bcbcbca
2211
12121111
ms
s
si
m
i
m
i
mi
i
c
cb
c
cb
c
cb
a
a1
2
12
2
1
11
1
1
This implies that each column vector of A lies in a space spanned by a single set of s vectors. Since s is the dimension of the row space of A, we get
dim(column space of A) dim(row space of A)
Ch04_68
By similar reasoning, we can show that
dim(row space of A) dim(column space of A)
Combining these two results we see that
dim(row space of A) = dim(column space of A),
proving the theorem.
DefinitionThe dimension of the row space and the column space of a matrix A is called the rank ( 秩 ) of A. The rank of A is denoted rank(A).
Ch04_69
Example 2Determine the rank of the matrix
852210321
A
Solution
The third row of A is a linear combination of the first two rows:(2, 5, 8) = 2(1, 2, 3) + (0, 1, 2)
Hence the three rows of A are linearly dependent. The rank of A must be less than 3. Since (1, 2, 3) is not a scalar multiple of (0, 1, 2), these two vectors are linearly independent. These vectors form a basis for the row space of A. Thus rank(A) = 2.
Ch04_70
Theorem 4.14The nonzero row vectors of a matrix A that is in reduced echelon form are a basis for the row space of A. The rank of A is the number of nonzero row vectors.
Proof
Let A be an m n matrix with nonzero row vectors of r1, …, rt. Consider the identity
Where k1, …, kt are scalars.
0rrr ttkkk 2211
The first nonzero element of r1 is 1. r1 is the only one of the row to have a nonzero number in this component. Thus, on adding the vectors we get a vector whose first component is k1. On equating this vector to zero, we get k1 = 0. The identity then reduced to
,,...,, 2211 ttkkk rrr
0rr ttkk 22
Ch04_71
The first nonzero element of r2 is 1, and it is the only of these remaining row vectors with a nonzero number in this component. Thus k2 = 0.
Similarly, k3, …, kt are all zero.
The vector r1, …, rt are therefore linearly independent. These vectors span the row space of A.
They thus form a basis for the row space of A. The dimension of the row space is t. The rank of A is t, the number of nonzero row vectors in A.
Ch04_72
Example 3
Find the rank of the matrix
This matrix is in reduced echelon form. There are three nonzero row vectors, namely (1, 2, 0, 0), (0, 0, 1, 0), and (0, 0, 0, 1). According to the previous theorem, these three vectors form a basis for the row space of A. Rank(A) = 3.
0000100001000021
A
Ch04_73
Theorem 4.15Let A and B be row equivalent matrices. Then A and B have the same the row space. rank(A) = rank(B).
Theorem 4.16
Let E be a reduced echelon form of a matrix A. The nonzero row vectors of E form a basis for the row space of A. The rank of A is the number of nonzero row vectors in E.
Note: If A … E, 且 E 是 reduced echelon form, then (a) 將 E 的每一非零列視為向量， 這些列形成一個 A 的列空間的 basis 。 (b) rank(A) = E 的非零列個數
Ch04_74
Example 4Find a basis for the row space of the following matrix A, and determine its rank.
511452321
A
Solution
Use elementary row operations to find a reduced echelon form of the matrix A. We get
000210701
210210321
511452321
The two vectors (1, 0, 7), (0, 1, 2) form a basis for the row space of A. Rank(A) = 2.
隨堂作業： 5(b), 12
Ch04_75
Example 5Find a basis for the column space of the following matrix A.
641
232
011
A
Solution
620
431
121tA
The column space of A becomes the row space of At. Let us find a basis for the row space of At.
000310501
620310121
620431121
310
,501
form a basis for the column space of A.
Ch04_76
Example 6Find a basis for the subspace V of R4 spanned by the vectors
(1, 2, 3, 4), (1, 1, 4, 2), (3, 4, 11, 8)Solution
000021100501
422021104321
8114324114321
(1, 0, 5, 0) and (0, 1, 1, 2) form a basis for the subspace V.
隨堂作業： 7(b)
8114324114321
ALet .
dim(V) = 2.
Ch04_77
Theorem 4.17Consider a system AX=B of m equations in n variables
(a) If the augmented matrix and the matrix of coefficients have the same rank r and r = n, the solution is unique.
(b) If the augmented matrix and the matrix of coefficients have the same rank r and r < n, there are many solutions.
(c) If the augmented matrix and the matrix of coefficients do not have the same rank, a solution does not exist.
隨堂作業： 10(a,b,c) 僅做 (i)(ii)
Ch04_78
Theorem 4.18Let A be an n n matrix. The following statements are equivalent.
(a) |A| 0 (A is nonsingular).
(b) A is invertible.
(c) A is row equivalent to In.
(d) The system of equations AX = B has a unique solution.
(e) rank(A) = n.
(f) The column vectors of A form a basis for Rn.
Ch04_79
Homework
Exercise 4.5:5, 7, 10, 12
Ch04_80
4.6 Orthonormal Vectors and Projections
DefinitionA set of vectors in a vector space V is said to be an orthogonal ( 正交 ) set if every pair of vectors in the set is orthogonal. The set is said to be an orthonormal set if it is orthogonal and each vector is a unit vector.
Ch04_81
Example 1
5
3 ,
5
4 0, ,
5
4 ,
5
3 0,),0 ,0 ,1(
Solution
(1) orthogonal: ;0,,0)0,0,1( 54
53
;0,,0)0,0,1( 53
54
;0,,0,,0 53
54
54
53
(2) unit vector:
10 ,0,
10 , 0,
1001)0,0,1(
222
222
222
53
54
53
54
54
53
54
53
Thus the set is thus an orthonormal set. 隨堂作業： 1,3(a)
Show that the set is an orthonormal set.
Ch04_82
Theorem 4.19An orthogonal set of nonzero vectors in a vector space is linearly independent.
Proof Let {v1, …, vm} be an orthogonal set of nonzero vectors in a vector space V. Let us examine the identity
c1v1 + c2v2 + … + cmvm = 0Let vi be the ith vector of the orthogonal set. Take the dot product of each side of the equation with vi and use the properties of the dot product. We get
Since the vectors v1, …, v2 are mutually orthogonal, vj‧vi = 0 unless j = i. Thus
Since vi is a nonzero, then vi‧vi 0. Thus ci = 0. Letting i = 1, …, m, we get c1 = 0, cm = 0, proving that the vectors are linearly independent.
0 iiic vv
0)(
2211
2211
immii
iimm
cccccc
vvvvvvv0vvvv
Ch04_83
DefinitionA basis is an orthogonal set is said to be an orthogonal basis. A basis that is an orthonormal set is said to be an orthonormal basis.
Standard Bases• R2: {(1, 0), (0, 1)}• R3: {(1, 0, 0), (0, 1, 0), (0, 0, 1)} orthonormal bases• Rn: {(1, …, 0), …, (0, …, 1)}
Theorem 4.20Let {u1, …, un} be an orthonormal basis for a vector space V. Let v be a vector in V. v can be written as a linearly combination of these basis vectors as follows:
nn uuvuuvuuvv )()()( 2211
Ch04_84
Example 2The following vectors u1, u2, and u3 form an orthonormal basis for R3. Express the vector v = (7, 5, 10) as a linear combination of these vectors.
5
3 ,
5
4 0, ,
5
4 ,
5
3 0, 0), 0, 1,( 321 uuu
Solution
105
3 ,
5
4 0,10) ,5 ,7(
55
4 ,
5
3 0,10) ,5 ,7(
70) 0, ,1(10) ,5 ,7(
3
2
1
uv
uv
uv
Thus
321 1057 uuuv
隨堂作業： 4
Ch04_85
Orthogonal MatricesAn orthogonal matrix is an invertible matrix that has the property
A1 = At
Theorem 4.21 (Orthogonal Matrix Theorem)
The following statements are equivalent.(a) A is orthogonal. (b) The column vectors of A form an orthonormal set.(c) The row vectors of A form an orthonormal set.
(a bc) A is orthogonal A1 = At AtA= I and AAt= I The column vectors of A form an orthonormal set, and the row vectors of A form an orthonormal set. 反之亦然
Proof
Ch04_86
Theorem 4.22If A is an orthogonal matrix, then(a) |A| = 1.(b) A1 is an orthogonal matrix.
Proof
(b) (A1)t (A1)= AAt= I A1 is an orthogonal matrix
(a) AAt= I |AAt| = |A||At| = |A||A1| = 1 |A| = 1
隨堂作業： 9
Ch04_87
Projection of One vector onto Another Vector
Figure 4.7
Let v and u be vectors in Rn with angel (0 ) between them.
||||
|||| |||| ||||
cos |||| cos
u
uv
uv
uvv
v
OBOA
OA : the projection of v onto u
uuu
uv
u
u
u
uv
)||||
)(||||
( OA
.0 then 2 / If : Note
uu
uv
.proj uuu
uvvu
So we define
Ch04_88
DefinitionThe projection ( 投影 ) of a vector v onto a nonzero vector u in Rn is denoted projuv and is defined by
uuuuv
vu proj
Figure 4.8O
Ch04_89
Example 3Determine the projection of the vector v = (6, 7) onto the vector u = (1, 4).
Solution
171614) (1,4) (1,342864) (1,7) (6,
uuuv
Thus
The projection of v onto u is (2, 8).
8) (2,4) ,1(1734
proj uuuuv
vu
隨堂作業： 14(b)
Ch04_90
Theorem 4.23The Gram-Schmidt Orthogonalization ProcessLet {v1, …, vn} be a basis for a vector space V. The set of vectors {u1, …, un} defined as follows is orthogonal. To obtain an orthonormal basis for V, normalize each of the vectors u1, …, un .
nnnn nvvvu
vvvu
vvuvu
uu
uu
u
11
21
1
projproj
projproj
proj
3333
222
11
Figure 4.9
Ch04_91
Example 4The set {(1, 2, 0, 3), (4, 0, 5, 8), (8, 1, 5, 6)} is linearly independent in R4. The vectors form a basis for a three-dimensional subspace V of R4. Construct an orthonormal basis for V.
Solution
Let v1 = (1, 2, 0, 3), v2 = (4, 0, 5, 8), v3 = (8, 1, 5, 6)}. Use the Gram-Schmidt process to construct an orthogonal set {u1, u2, u3} from these vectors.
2) 0, ,1 ,4()(
)(
)(
)(
projprojLet
2) 5, 4, 2,()()(
projLet
3) 0, 2, 1,(Let
222
231
11
133
3333
111
222222
11
21
1
uuu
uvu
uu
uvv
vvvu
uuuuv
vvvu
vu
uu
u
Ch04_92
The set {(1, 2, 0, 3), (2, 4, 5, 2), (4, 1, 0, 2)} is an orthogonal basis for V. Normalize them to get an orthonormal basis:
21)2(0142) 0, 1, (4,
725)4(22) 5, 4, (2,
1430213) 0, 2, (1,
2222
2222
2222
orthonormal basis for V:
212
0, ,211
,214
,72
,75
,74
,72
,143
0, ,142
,141
隨堂作業： 16(a)
Ch04_93
Projection of a Vector onto a Subspace
DefinitionLet W be a subspace of Rn, Let {u1, …, um} be an orthonormal basis for W. If v is a vector in Rn, the projection of v onto W is denoted projWv and is defined by
mmW uuvuuvuuvv )()()(proj 2211
Figure 4.11
Ch04_94
Theorem 4.24Let W be a subspace of Rn. Every vector v in Rn can be written uniquely in the form
v = w + w
where w is in W and w is orthogonal to W. The vectors w and w are
w = projWv and w = v – projWv
Figure 4.12
Ch04_95
Example 5Consider the vector v = (3, 2, 6) in R3. Let W be the subspace of R3 consisting of all vectors of the form (a, b, b). Decompose v into the sum of a vector that lies in W and a vector orthogonal to W.
Solution
We need an orthonormal basis for W. We can write an arbitrary vector of W as follows
(a, b, b) = a(1, 0,0) + b( 0, 1, 1)
The set {(1, 0, 0), (0, 1, 1)} spans W and is linearly independent. It forms a basis for W. The vectors are orthogonal. Normalize each vector to get an orthonormal basis {u1, u2} for W, where
21
,2
1 ,0 0), 0, 1,( 21 uu
Ch04_96
4) 4, (3,4) 4, (0,0) 0, ,3(
21
,2
1 0,
21
,2
1 0,6) 2, (3,0) 0, 0))(1, 0, (1,6) 2, 3,((
)()(proj 2211
uuvuuvvw W
2) ,2 (0,4) 4, (3,6) 2, (3,proj vvw W
and
We get
Thus the desired decomposition of v is
(3, 2, 6) = (3, 4, 4) + (0, -2, 2)
In this decomposition the vector (3, 4, 4) lies in W and the vector (0, -2, 2) is orthogonal to W.
隨堂作業： 21(a)
Ch04_97
Distance of a Point from a SubspaceThe distance of a point from a subspace is the distance of the point from its projection in the subspace.
xxx WWd proj) ,(
Figure 4.13
Ch04_98
Example 6Find the distance of the point x = (4, 1, 7) of R3 from the subspace W consisting of all vectors of the form (a, b, b).
Solution
The previous example tells us that the set {u1, u2} where
is an orthonormal basis for W. We compute projWx
21
,2
1 ,0 0), 0, 1,( 21 uu
3) 3, (4,3) 3, (0,0) 0, ,4(
7) 1, (4,0) 0, 0))(1, 0, (1,7) 1, 4,((
)()(proj
2
1 ,
2
1 0,
2
1 ,
2
1 0,
2211
uuxuuxxW
Thus, the distance from x to W is
32)4 4, ,0(
)3 ,3 ,4()7 1, ,4(proj
xx W
隨堂作業： 26
Ch04_99
Homework
Exercise 4.9:1, 3, 4, 6, 9, 14, 16, 21, 26