Chapter 4 Vector Spaces - scut.edu.cn · 2019-11-28 · 2 1-dimensional subspaces. Any subspace...
Transcript of Chapter 4 Vector Spaces - scut.edu.cn · 2019-11-28 · 2 1-dimensional subspaces. Any subspace...
SCUT, Liu Rui
Chapter 4 Vector Spaces
Liu Rui
School of Mathematics
South China University of Technology
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;