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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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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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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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Copyright Ch05B_18

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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