Transcript General Vector Spaces II - Professor Shaw’s Teaching and
MAC 2103
Module 9 General Vector Spaces II
1
Learning Objectives
Upon completing this module, you should be able to: 1.
2.
3.
Find the coordinate vector with respect to the standard basis for any vector in ℜ ⁿ.
Find the coordinate vector with respect to another basis. Determine the dimension of a vector space V from a basis for V. 4.
5.
6.
Find a basis for and the dimension of the null space of A, null(A).
a) b) Find a basis for and the dimension of the column space of A, col(A). Show that the non-leading columns of A are linearly dependent since they can be written as a linear combination of the leading columns of A.
Show that the leading columns of A are linearly independent and therefore form a basis for col(A).
Find a basis for and the dimension of the row space of A, row(A).
Rev.F09
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Rev.09
General Vector Spaces II
There are three major topics in this module: Coordinate Vectors Basis and Dimension Null Space, Column Space, and Row Space of a Matrix
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Quick Review In module 8, we have learned that if we let S = {
v 1
,
v 2
, … ,
v r
} be a finite set of non-zero vectors in a vector space V, the vector equation
k
1 r
v
1
k
2 r
v
2 ...
k r
r
v r
r
i
1
k i
r
v i
r 0 has at least one solution, namely the trivial solution , 0 = k 1 = k 2 = … = k r . If the only solution is the trivial solution, then S is a linearly independent set. Otherwise, S is a linearly dependent set. If every vector in the vector space V can be expressed as a linear combination of the vectors in S, then S is the spanning set of the vector space V. If S is a linearly independent set, then S is a basis for V and span(S) = V.
Rev.F09
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What is the Coordinate Vector with Respect to the Standard Basis for any Vector?
The set of standard basis vectors in ℜ ⁿ is B = {
e 1
,
e 2
, … ,
e n
}. If
v
∈ ℜ ⁿ, then
T v
v
1
v
2 ...
v n
v
1 ˆ 1
v
2
e
ˆ 2 ...
v n e
ˆ
n
and has components
v
1
,
v
2
,...,
v n
.
The coordinate vector
v B
has the coefficients from the linear combination of the basis vectors as its components. So,
T v B
v
1
v
2 ...
v n
.
A better name might be coefficient vector, but it is not Rev.F09
used. So,
v
the standard basis,
v
r
v B
.
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What is the Coordinate Vector with Respect to Another Basis?
Example:
Let B 1 = {
v 1
,
v 2
}, with To show B 1 2
i
1
c i
r
v i
c
1 r
v
1
v
1 1 2 , r
v
2 1 1 .
is a basis, we solve
c
2 r
v
2 [ r
v
1 r
v
2 ] r
c
A
r
c
If the only solution is
c
c
1
c
2 0 , 0 r 0 1 2 1 1
c
1
c
2 r 0.
then
v 1
,
v 2
are linearly independent, and B 1 = {
v 1
,
v 2
} is a linearly independent set and a basis for ℜ ². Rev.F09
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What is the Coordinate Vector with Respect to Another Basis? (Cont.) The
det(
A
)
1 2
1
1
1
0
and A -1 exists, so the only solution to A
c
=
0
is A -1 A
c
= I 2
c
= A -1
0
=
0
; so,
c
=
0
. Thus, B 1 = {
v 1
,
v 2
}, is a basis for ℜ ² and not the standard basis.
Rev.F09
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Let What is the Coordinate Vector with Respect to
v
c
1 1 3 1 2 Another Basis? (Cont.)
c
1 r
v
1
c
2 r
v
2 [ r
v
1 r
v
2 ]
c
2 1 1 1 2 1 1
c
1
c
2
c
1
c
2
A
r
c
r
v
We want to solve A
c
=
v
for
c
, the vector of coefficients, which is the coordinate vector
v B
1 .
v
] as follows:
A
| r
v
1 2 1 1 1 3
r
2
r
1 2
r
1 1 0 1 1 1 1 Thus, Rev.F09
c
2 1,
c
1 2 and 1 3 (2) http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. 1 2 (1) 1 1 .
8
What is the Coordinate Vector with Respect to Another Basis? (Cont.) So, the c i that are the components of the coordinate vector
v B
1 are the coefficients in the linear combination of the basis vectors for
v
.
Thus, the coordinate vector with respect to B 1 is
v B
1 ( r
v
)
B
1
c c
1 2 2 1 (
c
1 ,
c
2 ) (2,1).
Rev.F09
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What is a Basis for a Vector Space, and What is the Dimension of a Vector Space?
Let
v
1 , r
v
2 ,..., r
v n
V
.
the vector space V Then
S
{ r
v
1 , r
v
2 ,..., r
v n
} is a basis for if both of the following conditions hold: 1. S is a set of linearly independent vectors or it is a linearly independent set, and 2. The vectors in S can span the vector space V. This means that the
v
1 , r
v
2 ,..., r
v n
The dimension of a vector space is the number of vectors in any basis for the vector space. dim(V) = n. The vector space V could have infinitely many bases, for V ≠ span({
0
}).
Rev.F09
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How to Find a Basis for and the Dimension of the Null Space of A, Null(A)? Example: Find a basis for and the dimension of the null space of A, null(A), which is the solution space of the homogeneous system:
A
r
x
r 0 2
x
1
x
1 2
x
2
x
2
x
2
x
3 3
x
1
x
2 2
x
3
x
5 3
x
4
x
x
5 5 0 0 0
x
3
x
4
x
5 0.
We shall use Gauss Elimination to obtain a row echelon form.
r r r r
1 2 3 4 Rev.F09
2 1 1 0 2 1 1 0 1 2 2 1 0 0 1 3 1 1 1 1 0 0 0 0
A
| r 0 http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. r
a
1 r
a
2 r
a
3 r
a
4 11 r
a
5 | r 0
How to Find a Basis for and the Dimension of the Null Space of A, Null(A)?(Cont.) 1) 3) 1 2
r
1
r
2
r
3
r
4 2)
r
1
r
1
r
2
r
1
r
3
r
4
r
1 2 3
r
2
r
3
r
4 1 1 1 0 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 0 1 0 0 0 Rev.F09
1 2 1 3 2 1 1 2 2 2 1 1 2 3 2 3 2 1 0 3 0 1 0 3 0 1 0 2 0 1 1 2 1 1 1 2 1 1 1 2 3 2 3 2 1 1 3 2 0 0 0 0 0 0 0 0 0 0 0 0 4) 5) 6) 3 2
r
2
r
2
r
2
r
3
r
4
r r r
1 3 1 2 4
r
3 3
r r r r
3
r
1 1 2 3
r
4 http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 2 1 0 0 1 2 1 0 0 1 2 1 0 0 0 2 1 3 0 2 3 3 0 2 1 0 12 1 0 0 1 2 1 0 0 1 2 1 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0
How to Find a Basis for and the Dimension of the Null Space of A, Null(A)?(Cont.) 1 0 0 0 1 0 0 0 1 0 0 1 2 0 2 1 0 1 2 1 0 0 0 0 0 0
G
| r 0 G has red leading 1’s in column 1, column 3, and column 4. These are the three leading columns of G. G is a row-echelon form of A. These columns are linearly independent and correspond to the linearly independent
a
1 , r
a
3 , r
a
4 , which form a basis columns of A, for the column space of A, col(A). The non-leading columns of G are column 2 and column 5 which correspond to linearly dependent columns of A and give us the free variables, x 2 = s and x 5 homogeneous system, A
x
=
0
.
Rev.F09
= t in our solution of the http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. 13
How to Find a Basis for and the Dimension of the Null Space of A, Null(A)?(Cont.) 1 0 0 0 1 0 0 0 1 2 1 0 0 0 2 1 0 1 2 1 0 0 0 0 0 0
G
| r 0 We know x 2 = s and x 5 = t.
Use back-substitution to find the solution
x
.
x
5
x
2
t
,
x
4
s
,
x
1 0,
x
3
x
2 1 2 2
x
4
x
3 1 2
x
5
x
5
t
,
s
t
All of the components of
x
are in terms of s and t.
Rev.F09
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How to Find a Basis for and
x
the Dimension of the Null Space of A, Null(A)?(Cont.)
x
1
x
2
x
3
x
4
x
5
s
t s
t t
0
s
1 1 0 0 0
t
1 0 1 0 1
s
r
v
1
t
r
v
2 The solution The
x
of A solution space
x
=
0
is a linear combination of
v 1
and
v 2
is the span({
v 1
,
v 2
}), the set {
v 1
,
v 2
} is . linearly independent (since
v 1
is not a multiple of
v 2
) and is a basis for the solution space of the homogeneous system or the null space of A , null(A). So, null(A) = span({
v 1
,
v 2
}). Null(A) is a subspace of ℜ ⁵ and has a dimension of 2, dim(null(A))=2. Thus,
x
s
r
v
1
t
r
v
2 for some s and t iff
x
is in the null(A). Rev.F09
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How to Find a Basis for and the Dimension of the 2 1 1 0 1 0 0 0 2 1 1 0 1 0 0 0 1 2 2 1 0 0 1 1 2 Column Space of A, Col(A)?
0 3 1 0 0 1 0 2 1 2 1 0 0 1 1 1 1 0 0 0 0 0 0 0 0
G
A
| | r 0 r 0 r
a
1 r
a
2 r
a
3 r
a
4 r
a
5 | r 0 By definition, the column space of A, col(A) = span({
a 1
,
a 2 , a 3 a 4 , a 5
}), but not all of the vectors in the set are linearly independent. The linearly independent columns of A correspond to the leading columns of G ; hence, col(A) = span({
a 1
,
a 3 , a 4
}), and {
a 1
,
a 3 , a 4
} is a basis for col(A). Then dim(col(A))=3. We will prove these statements for A in the next few slides.
Rev.F09
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How to Find a Basis for and the Dimension of the Column Space of A, Col(A)? (Cont.)
G
| r 0 1 0 0 0 1 0 0 0 1 2 1 0 0 0 2 1 0 1 0 0 1 2 0 0 0 0 1 2
r
2
r
1
r
2
r
3
r
4 1 0 0 0 1 0 0 0 0 1 0 0 1 2 1 0 1 1 0 0
r
3 2
r
3
r
1
r
2
r
3
r
4 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 r
v
1 r
v
2 r
v
3 r
v
4 r
v
5 | r 0
J
| r 0 The last matrix is the reduced row-echelon form for [A|
0
]. We can see that
v 2
=(1)
v 1
+(0)
v 3
+(0)
v 4
=
v 1
or
v 2
=
v 1
.
v 5
=(1)
v 1
+(1)
v 3
+(0)
v 4
=
v 1
+
v 3
or
v 5
=
v 1
+
v 3
. Thus,
v 2
and
v 5
linearly dependent columns in the span of {
v 1
,
v 3
,
v 4
}.
are Rev.F09
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How to Find a Basis for and the Dimension of the Column Space of A, Col(A)?(Cont.) 2 1 1 0 2 1 1 0 1 2 2 1 0 3 0 1 1 1 1 1 0 0 0 0
A
| r 0 r
a
1 r
a
2 r
a
3 r
a
4 r
a
5 | r 0 The linear dependencies will hold for the columns of A. We can see that
a 2
=
a 1
, and
a 5
=
a 1
+
a 3
. Thus,
a 2
and
a 5
are linearly dependent columns in the span{
a 1
,
a 3
,
a 4
} = col(A). Rev.F09
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How to Find a Basis for and the Dimension of the 1 0 0 0 1 0 0 0 Column Space of A, Col(A)? (Cont.) 0 1 0 0 0 0 1 0 1 1 0 0 0 0 0 0 r
v
1 r
v
2 r
v
3 r
v
4 r
v
5 | r 0 [
J
| r 0] J is the reduced echelon form of A. The leading columns
v 1
,
v 3
,
v 4
are the linearly independent columns of J, since they are standard basis vectors in ℜ 4 .
v 1
=
e 1
,
v 3
=
e 2
,
v 4
=
e 3
.
c
1
e
ˆ 1
c
2
e
ˆ 2
c
3
e
ˆ 3
c
1 1 0 0 0
c
2 0 1 0 0
c
3 0 0 1 0
c c c
0 1 2 3 0 0 0 0
c
1
c
2 Rev.F09
c
3 0 which proves linear independence.
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How to Find a Basis for and the Dimension of the Column Space of A, Col(A)? (Cont.) set. Let
c
1 r
a
1
c
2 r
a
3
c
3 r
a
4
a 1
, r 0
a 3
,
a 4
} is a linearly independent and let E be the product of all elementary matrices, such that EA=J. Then,
EA
E
r
a
1
E
r
a
2
E
r
a
3
E
r
a
4
E
r
a
5
J
r
v
1 r
v
2 r
v
3 r
v
4 with leading columns E
a 1
=
v 1
=
e 1
, E
a 3
=
v 3
=
e 2
, E
a 4
=
v 4
=
e 3
, as seen from the previous slide. Multiplying by E gives us
E
(
c
1 r
a
1
c
2 r
a
3
c
3 r
a
4 )
c
2
E
r
a
3
c
3
E
r
a
4
c
1
e
ˆ 1
c
2
e
ˆ 2
c
3
e
ˆ 3 r
c c
1
E
r 0.
r
a
1
c
=
0
is the unique solution since E is invertible. Therefore, {
a 1
,
a 3
,
a 4
} is a linearly independent set and a basis for the col(A). Then, dim(col(A)) = 3.
r
v
5 Rev.F09
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How to Find a Basis for and the Dimension of the Row Space of A, Row(A)? 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 1 0 0 r
w
1 r
w
2 r
w
3 r
w
4 The nonzero row vectors in the matrix J are linearly independent. The row vectors
w 1
,
w 2 , w 3
form a basis for the row space of J. Likewise, the nonzero row vectors in the matrix G are linearly independent and those three row vectors form a basis for row(G).
Rev.F09
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How to Find a Basis for and the Dimension of the Row Space of A, Row(A)? (Cont.) Elementary row operations are linear operators from ℜ ⁵ into ℜ ⁵ and the new rows are linear combinations of the original rows. So, row(A) = row(G) = row(J), and the span({
w 1
,
w 2
, }) = row(A). Then dim(row(A))=3.
w 3
In A, 2
w 1
-
w 2
=
r 1
, -
w 1
+2
w 2
-3
w 3
=
r 2
,
w 1
-2
w 2
=
r 3
, and
w 2
+
w 3
=
r 4
. This proves row(J) = row(A). A basis for the row(A) is {
w 1
,
w 2
,
w 3
} = {[1 1 0 0 1], [0 0 1 0 1], [0 0 0 1 0]}.
The row(A) = col(A T ) since the rows of A are the columns of A T . If we only find a basis for row(A), then we can find a basis for col(A T ) and switch the column vectors back to row vectors (see Example 8 on page 274).
Rev.F09
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What have we learned?
We have learned to : 1.
2.
3.
Find the coordinate vector with respect to the standard basis for any vector in ℜ ⁿ.
Find the coordinate vector with respect to another basis. Determine the dimension of a vector space V from a basis for V. 4.
5.
6.
Find a basis for and the dimension of the null space of A, null(A).
a) b) Find a basis for and the dimension of the column space of A, col(A). Show that the non-leading columns of A are linearly dependent since they can be written as a linear combination of the leading columns of A.
Show that the leading columns of A are linearly independent and therefore form a basis for col(A).
Find a basis for and the dimension of the row space of A, row(A).
Rev.F09
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Credit
Some of these slides have been adapted/modified in part/whole from the following textbook: • Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition Rev.F09
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