SCUT, Liu Rui

Chapter 4 Vector Spaces

Liu Rui

School of Mathematics

South China University of Technology

scliurui@scut.edu.cn

2019-11-29

SCUT, Liu Rui

1 § 4.5 The Dimension of A Vector Space

2 § 4.6 Rank

3 § 4.7 Change of basis

SCUT, Liu Rui

The Dimension of A Vector Space

Definition

The dimension of a nonzero vector space V , denoted by dim V , is

the number of vectors in any basis for V .

SCUT, Liu Rui

The Dimension of A Vector Space

Proposition (1)

1 If the vector space V is spanned by a finite set, then the di-

mension of V is finite, V is finite-dimensional.

2 The dimension of the zero vector space {~0} is defined to be

zero. The zero vector space has no basis.

3 If the vector space V is not spanned by a finite set, V is

infinite-dimensional.

Example for the infinite-dimensional vector space: R∞, con-

tinuous function space C(R)

SCUT, Liu Rui

The Dimension of A Vector Space

Proposition (2)

1 If a vector space V has a basis B = {~b1, ...,~bn}, then any

set in V containing more than n vectors must be linearly de-

pendent.

2 If a vector space V has a basis B = {~b1, ...,~bn}, then any

other bases of V must consist of exactly n vectors.

SCUT, Liu Rui

The Dimension of A Vector Space

Example (1)

1 dim Rn = n

2 dim Pn = n + 1

3 dim Mn×n = n2

SCUT, Liu Rui

The Dimension of A Vector Space

Example (2)

It is obvious that ~v1 =

3

6

2

, ~v2 =

−10

1

are linearly inde-

pendent. Therefore, for the vector space

H = span{~v1, ~v2},

the dimension dim H = 2.

SCUT, Liu Rui

The Dimension of A Vector Space

Example (3)

Find the dimension of the vector space

H =

a− 3b + 6c

5a + 4d

b− 2c− d

5d

: a, b, c, d ∈ R

.

SCUT, Liu Rui

The Dimension of A Vector Space

Solution:

Obviously,a− 3b + 6c

5a + 4d

b− 2c− d

5d

= a

1

5

0

0

+ b

−30

1

0

+ c

6

0

−20

+ d

0

4

−15

.

SCUT, Liu Rui

The Dimension of A Vector Space

Then

H = span

1

5

0

0

,

−30

1

0

,

6

0

−20

,

0

4

−15

.

It is clear that ~v3 is a multiple of ~v2, and ~v1, ~v2, ~v4 are linearly

independent. Therefore, {~v1, ~v2, ~v4} is a basis for H.

dim H = 3.

�

SCUT, Liu Rui

The Dimension of A Vector Space

Example (4)

The subspaces of R3 can be classified by dimension.

1 0-dimensional subspaces. Only the zero subspace {~0}.

2 1-dimensional subspaces. Any subspace spanned by a single

nonzero vector, that is, any line through the origin.

3 2-dimensional subspaces. Any subspaces spanned by two lin-

early independent vectors, that is, any plane through the origin.

4 3-dimensional subspace. R3, any three linearly independent

vectors in R3 is a basis.

SCUT, Liu Rui

The Dimension of A Vector Space

SCUT, Liu Rui

The Dimension of A Vector Space

Proposition (3)

Let V be a p-dimensional vector space, p ≥ 1.

1 For any set B = {~v1, ~v2, ..., ~vp}, if B is a linearly independent

set, then B is automatically a basis for V .

2 For any set B = {~v1, ~v2, ..., ~vp}, if V =

span{~v1, ~v2, ..., ~vp}, then B is automatically a basis

for V .

SCUT, Liu Rui

The Dimension of A Vector Space

Proposition (4)

For a matrix A.

1 The dimension of Nul A is the number of free variables in the

equation A~x = ~0.

2 The dimension of Col A is the number of leading entries in an

echelon form (or reduced echelon form) of A.

SCUT, Liu Rui

The Dimension of A Vector Space

Determine if each statement is true or false (P245, 19)

SCUT, Liu Rui

The Dimension of A Vector Space

a. True

b. False

c. False

d. False

e. True

SCUT, Liu Rui

The Dimension of A Vector Space

Determine if each statement is true or false (P261, 20)

SCUT, Liu Rui

The Dimension of A Vector Space

a. False

b. False

c. False

d. False

e. True

SCUT, Liu Rui

Rank

Definition (Rank of a matrix)

The rank of a matrix A, denoted by rank A, is the dimension of the

column space of A.

B =

1 0 −3 5 0

0 1 2 −1 0

0 0 0 0 1

0 0 0 0 0

The basis of Col B is {~b1, ~b2, ~b5}.

rank A = dim (Col A) = 3

SCUT, Liu Rui

Rank

For a matrix A, the spanned space by the rows of A is called

the row space.

Example

Find bases for the row space, the column space.

SCUT, Liu Rui

Rank

Solution:

Basis for the row space

{Row 1, Row 2, Row 3} ⊂ B

Basis for the column space

{Column 1, Column 2, Column 4} ⊂ A

SCUT, Liu Rui

Rank

Proposition

For an m× n matrix A.

1 If n ≤ m, then rank A ≤ n.

2 If m ≤ n, then rank A ≤ m.

3 The dimension of the row space are equal to the rank of AT .

4 rank A + dimNul A = n (no matter n ≤ m or n > m).

5 The dimensions of the row space and the column space of A

are the same, even if A is not a square matrix.

SCUT, Liu Rui

Rank

Example

1 If A is a 7× 9 matrix with a two-dimensional null space, what

is the rank of A?

2 Could a 6× 9 matrix have a two-dimensional null space?

1 rank A = 7

2 No, otherwise rank A = 7 > 6

SCUT, Liu Rui

Rank

Example

1 If A is a 7× 9 matrix with a two-dimensional null space, what

is the rank of A?

2 Could a 6× 9 matrix have a two-dimensional null space?

1 rank A = 7

2 No, otherwise rank A = 7 > 6

SCUT, Liu Rui

Rank

Example

1 If A is a 7× 9 matrix with a two-dimensional null space, what

is the rank of A?

2 Could a 6× 9 matrix have a two-dimensional null space?

1 rank A = 7

2 No, otherwise rank A = 7 > 6

SCUT, Liu Rui

Rank and the invertible matrix theorem

Rank and its propositions about the invertible matrices:

Theorem (The invertible matrix theorem)

Let A be an n × n matrix. Then the following statements are

equivalent

1 A is an invertible matrix.

2 The columns of A form a basis of Rn.

3 Col A = Rn.

4 dim Col A = dim Row A = rank A = n.

5 Nul A = {~0}.

6 dim Nul A = 0.

SCUT, Liu Rui

Homework

Homework:

Section 4.5 p. 245: 6, 12, 21, 23;

Section 4.6 p. 252-253: 4, 27;

SCUT, Liu Rui

Change of basis

Example

Consider two bases B = {~b1,~b2} and C = {~c1, ~c2} for a vector

space V , such that

~b1 = 4~c1 + ~c2, ~b2 = −6~c1 + ~c2.

For a given vector ~x ∈ V , if

~x = 3~b1 +~b2,

find [~x]B and [~x]C .

SCUT, Liu Rui

Change of basis

Solution:

Firstly, [~x]B =

3

1

.

Secondly, since

[~x]C = [3~b1 +~b2]C

= 3[~b1]C + [~b2]C

The coordinate of 3~b1 + ~b2 under basis C is equivalent to

3[~b1]C + [~b2]C. Why?

SCUT, Liu Rui

Change of basis

Notice that [~b1]C =

4

1

and [~b2]C =

−61

.

Therefore,

[~x]C = 3[~b1]C + [~b2]C

= 3

4

1

+

−61

=

6

4

SCUT, Liu Rui

Change of basis

Solution 2:

[~x]C = 3[~b1]C + [~b2]C

=([~b1]C, [~b2]C

) 3

1

=

4 −61 1

3

1

=

6

4

SCUT, Liu Rui

Change of basis

A vector ~x under different bases:

[~x under B]

[~x under C]

SCUT, Liu Rui

Change of basis

4 −61 1

3

1

=

6

4

PB→C [~x]B [~x]C

PB→C is called the change-of-coordinates matrix from basis

B to basis C.

SCUT, Liu Rui

Change-of-basis theorem

Theorem (Change of basis)

Let B = {~b1, ...,~bn} and C = {~c1, ..., ~cn} be two bases of a

vector space V . Then there is a unique n× n matrix PB→C such

that

[~x]C = PB→C[~x]B.

The columns of PB→C are the C-coordinates of the vectors in the

basis B. That is

PB→C =([~b1]C [~b2]C ... [~bn]C

)

SCUT, Liu Rui

Change-of-basis theorem

The geometric meaning of PB→C (the change-of-coordinates

matrix form B to C):

SCUT, Liu Rui

Change-of-basis theorem

Proposition (1)

1 The change-of-coordinates matrix PB→C is invertible.

2 Left multiplication by PB→C transforms B-coordinates into

C-coordinates.

[~x]C = PB→C[~x]B.

3 Left multiplication by P−1B→C transforms C-coordinates into

B-coordinates.

[~x]B = P−1B→C[~x]C.

4 P−1B→C = PC→B

SCUT, Liu Rui

Change-of-basis theorem in Rn

Proposition (2. Change of basis in Rn)

If B = {~b1, ...,~bn} is a basis of Rn, and E = {~e1, ..., ~en} is the

standard basis, then

1 [~b1]E = ~b1.

2 PB→E =(~b1, ..., ~bn

)

SCUT, Liu Rui

Change of basis in Rn

Example

Let

B = {~b1,~b2} =

−9

1

,

−5−1

,

C = {~c1, ~c2} =

1

−4

,

3

−5

.

Find the change-of-coordinates matrix PB→C form B to C.

SCUT, Liu Rui

Change of basis in Rn

Solution:

The matrix PB→C involves the C-coordinate vectors of ~b1 and

~b2. Assume [~b1]C =

x1

x2

and [~b2]C =

y1

y2

. Then,

x1~c1 + x2~c2 = ~b1

y1~c1 + y2~c2 = ~b2

SCUT, Liu Rui

Change of basis in Rn

To solve both systems above, we augment the coefficient ma-

trices

Thus [~b1]C =

6

−5

and [~b2]C =

4

−3

.

The change-of-coordinates matrix is PB→C =

6 4

−5 −3

.

SCUT, Liu Rui

Change of basis in R2

Algorithm to find the change-of-coordinates matrix: Suppose

B ={~b1,~b2

}, C = {~c1, ~c2} .

are two basis of R2. Then the algorithm to find the change-

of-coordinates matrix from B to C is

This algorithm also works for finding the change-of-coordinates

matrix between any two bases in Rn.

SCUT, Liu Rui

Change of basis in Rn

Exercise:

Let

B = {~b1,~b2} =

7

5

,

−3−1

,

C = {~c1, ~c2} =

1

−5

,

−22

.

Find the change-of-coordinates matrix PB→C form B to C.

Find the change-of-coordinates matrix PC→B form C to B.

SCUT, Liu Rui

Homework

Homework:

Section 4.7 p. 259: 9, 10, 13, 14;