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Chapter 4(C)General Vector Spaces(II)
4.6 ~ 4.9
Gareth Williams
J & B,
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Copyright Ch05B_2
Part C
4.6 Linear Dependence and Independence
4.7 Basic and Dimension
4.8 Rank of a Matrix
4.9 Orthonormal Vector and Projections in
Rn
Homework 4
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4.6 Linear Dependence andIndependence
(4, -1, 0) = 2*(2, 1, 3)-3*(0, 1, 2)
(2, 1, 3) = 0.5*(4, -1, 0)+1.5*(0, 1, 2)
(4, -1, 0)2* (2, 1, 3) + 3* (0, 1, 2) = (0, 0, 0)
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Copyright Ch05B_4
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.
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Example 1
Show that the set {(1, 2, 3), (-2, 1, 1), (8, 6, 10)} is linear
dependent in R3.
Solution
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Example 2
Show that the set {(3, -2, 2), (3, -1, 4), (1, 0, 5)} is linear
independent in R3.
Solution
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Example 3Consider the functionsf(x) =x2 + 1, g(x) = 3x1, 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
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Example 3
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Theorem 4.8
A set consisting of two or more vectors space is linearly
dependent if and only if it is possible to express one of the vectorsas a linearly combination of the other vectors.
Proof
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Theorem 4.8
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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 4.21 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 on a line {v1, v2, v3} linearly independent;vectors do not lie on a line
Figure 4.22 Linear dependence and independence of {v1, v2, v3} in R3.
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Theorem 4.9
Let V be a vector space. Any set of vectors in V that contains the
zero is linearly dependent.
Proof
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Theorem 4.10
Let the set{v1, , vm}be linearly dependent in a vector space V.
Any set of vectors in V that contains these vectors will also belinearly dependent.
Proof
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Example 4
Let the set {v1, v2} be linearly independent. Prove that {v1 + v2,
v1v2} is also linearly independent.
Solution
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4.7 Bases and Dimension
DefinitionA finite set of vectors {v1, , vm} is called a basis for a vector
space Vif the set spans Vand is linearly independent.
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Copyright Ch05B_17
DefinitionThe set ofnvectors {(1, 0, , 0), (0, 1, , 0), , (0, , 1)}
is a basis for Rn. This Basis is called the standard basis for Rn.
Standard Basis
Proof
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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
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Example 1
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Example 2
Show that {f, g, h }, wheref(x) =x2 + 1, g(x) = 3x1, and h(x)
=4x + 1 is a basis for P2.
Solution
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Example 2
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Substituting for w1, , wm into Equation (1) we get
Rearranging, we get1 11 1 12 2 1 1 1 2 2( ) ( )n n m m m mn nc a a a c a a a v v v v v v 0
1 11 2 21 1 1 1 1 2 2( ) ( )m m n n m mn nc a c a c a c a c a c a v v 0
Since v1, , vn are linear independent, this identity can be
satisfied only if the coefficients are all zero. Thus
0
0
2211
1221111
mmnnn
mm
cacaca
cacaca
Thus finding cs that satisfy Equation (1) reduces to finding
solutions to this system ofn equations in m variables. Since m >
n, the number of variables is greater than the number ofequations.
We know that such a system of homogeneous equations has
many solutions. These are therefore nonzero ofcs that satisfy
Equation (1). Thus the set {w1, , wm} is linearly dependent.
Theorem 4.11
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Definition
If a vector space Vhas a basis consisting ofn vectors, then thedimension ofVis said to be n. We write dim(V) for the
dimension ofV.
finite dimension
infinite dimension
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Example 3
Consider the set {{1, 2, 3), (-2, 4, 1)} of vectors in R3. These
vectors generate a subspace VofR3 consisting of all vectors ofthe 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 Vis, in fact, a plane through the origin.
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Theorem 4.13
(a) The origin is a subspace ofR3. The dimension of this
subspace is defined to be zero.
(b) The one-dimensional subspaces ofR3 are lines through the
origin.
(c) The two-dimensional subspaces ofR3 are planes through the
origin.
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Figure 4.23 One and two-dimensional subspaces ofR3
Theorem 4.13
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Proof
(a) Let Vbe the set {(0, 0, 0)}, consisting of a single elements,the zero vector ofR3. 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 ofR3. The dimension of this subspaces is definedto be zero.
(b) Let v be a basis for a one-dimensional subspace VofR3.
Every vector in Vis 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 VofR
3.
Every vector in Vis of the form c1v1 + c2v2. Vis thus a plane
through the origin.
Theorem 4.13
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Theorem 4.14
Let {v1, , vn} be a basis for a vector space V. Then each vector
in Vcan be expressed uniquely as a linear combination of thesevectors.
Proof
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Theorem 4.15
Let Vbe a vector space of dimension n.
(a) If S= {v1, , vn} is a set ofn linearly independent vectorsin V, then Sis a basis for V.
(b) IfS= {v1, , vn} is a set ofn vectors that spans V, then Sis
a basis for V.
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Example 4Prove that the set {(1, 3, -1), (2, 1, 0), (4, 2, 1)} is a basis for R3.
Solution
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Theorem 4.16
Let Vbe a vector space of dimension n. Let {v1, , vm} be a set
ofm linearly independent vectors in V, where mn. Thenthere exist vectors vm+1,,vn such that {v1, , vm, vm+1, ,
vn } is a basis for V.
Proof
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Example 5
(a) The vectors (1, 2), (-1, 3), (5, 2) are linearly independent in R2.
False : The dimension ofR2 is two. Thus any three vectors
are linearly dependent.
False: The three vectors are linearly dependent. Thus theycannot span a three-dimensional space.
Solution
State (with a brief explanation) whether the following
statements are true or false.
(b) The vectors (1, 0, 0), (0, 2, 0), (1, 2, 0) span R3.
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False: The two vectors must be linearly independent.
(c){(1, 0, 2), (0, 1, -3)} is a basis for the subspace ofR3
consisting of vectors of the form (a, b, 2a3b).
Solution
True: The vectors span the subspace since
(a, b, 2a3b) = a(1, 0, 2) + b(0, 1, -3)
The vectors are also linearly independent since they
are not colinear.
Example 5
(d) Any set of two vectors can be used to generate a two-
dimensional subspace ofR3.
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Space of Matrices
Consider the vector spaceM22 of 2x2 matrices.
The following matrices spanM22.
1 0 0 1 0 0 0 00 0 0 0 1 0 0 1
a b a b c d c d
1 2 3 4
1 2
1 2 3 4
3 4
They are also linearly indenpent since
1 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 1 0 0
0 0, implying that 0, 0, 0, 0
0 0
c c c c
c cc c c c
c c
1 0 0 1 0 0 0 0, , ,
0 0 0 0 1 0 0 1
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Space of Polynomials
Consider the vector space P2 of polynomial of degree 2.
The functionx2,x, 1 span P2.
2 2( ) ( ) (1)ax bx c a x b x c
2
2
1 2 3 1 2 3
2
2
, ,1 are linearly independent since
( ) ( ) (1) 0 implies that 0, 0, 0.
, ,1 is a basis for .
x x
c x c x c c c c
x x P
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Space of Cn
(1, 0) and (0, 1) are also linearly independent in C2. Thus {(1,
0), (0, 1)} is a basis for C2, and the dimension of C2 is 2.
Consider the complex vector space C2. The vector (1, 0) and
(0, 1) span C
2
since an arbitrary (a+bi, c+d
i) can be written:
(a+bi, c+di) = (a+bi) (1, 0) + (c+di) (0, 1)
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4.8 Rank of a Matrix
Definition
LetA be an mn matrix. The rows ofA may be viewed as rowvectors 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 ofA, and the column vectors will span a
subspace ofRm called the column space ofA.
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Example 1
Consider the matrix
The row vectors ofA are
0145
61432121
These vectors span a subspace ofR4 called the row space ofA.
The column vectors ofA are
These vectors span a subspace ofR3 called the column space
ofA.
0
6
2
1
1
1
4
4
2
5
3
1
4321cccc
)0,1,4,5(),6,1,4,3(),2,1,2,1( 321 rrr
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Theorem 4.17
The row space and the column space of a matrixA have the
same dimension.
Proof
Let u1, , um be the row vectors ofA. 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 thejth vector of this set be
)...,,,( 21 iniii aaau
)...,,,(21 jnjjj bbbv
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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
c
b
c
c
b
c
c
b
a
a
1
2
12
2
1
11
1
1
Each of the row vectors ofA is a linear combination ofv1, ,
vs. Let
smsmmm
ss
ccc
ccc
vvvu
vvvu 1
2211
1212111
Theorem 4.17
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Definition
The dimension of the row space and the column space of a matrixA is called the rank ofA. The rank ofA is denoted rank(A).
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Example 2Determine the rank of the matrix
852
210
321
A
Solution
Th
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Theorem 4.18
The nonzero row vectors of a matrixA that is in reduced echelon
form are a basis for the row space ofA. The rank ofA is thenumber of nonzero row vectors.
Proof
E l 3
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Example 3
Find the rank of the matrix
0000
1000
0100
0021
A
Th 4 19
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LetA andB be row equivalent matrices. ThenA andB have the
same the row space. rank(A) = rank(B).
Theorem 4.19
Th 4 20
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Theorem 4.20
LetEbe a reduced echelon form of a matrixA. The nonzero row
vectors ofEform a basis for the row space ofA. The rank ofAis the number of nonzero row vectors inE.
E l 4
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Example 4Find a basis for the row space of the following matrix A, and
determine its rank.
511
452321
A
Solution
E l 5
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Example 5Find a basis for the column space of the following matrix A.
641
232
011
A
Solution
E l 6
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Example 6Find a basis for the subspace VofR4 spanned by the vectors
(1, 2, 3, 4), (-1, -1, -4, -2), (3, 4, 11, 8)
Solution
Th 4 21
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Theorem 4.21
Consider a system ofm equations in n variables
(a) If the augmented matrix and the matrix of coefficients havethe same rankrand r= n, the solution is unique.
(b) If the augmented matrix and the matrix of coefficients have
the same rankrand r< n, there are many solution.
(c) If the augmented matrix and the matrix of coefficients do nothave the same rank, a solution does not exist.
E l 7
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Example 7Consider the following system of linear equations:
1 2 3
1 2 3
1 2 3
2
2 3 3
2 6
x x x
x x x
x x x
Solution
Th 4 22
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Theorem 4.22
LetA be an nn matrix. The following statements are
equivalent.
(a) |A| 0 (A is nonsingular.)
(b) A is invertible.
(c) A is row equivalent toIn
.
(d) rank(A) = n.
(e) The column vectors ofA form a basis for Rn.
(f) The system of equations AX=B has a unique solution.
E ample 8
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Example 8Consider the matrixA:
Compute the reduced echelon from ofA. Discuss theimplications of the answer.
Solution
1 1 22 3 5
1 3 5
A
Example 8
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Example 8
Solution
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