4.1 vector spaces and subspaces 4.2 null spaces, column spaces, and linear transformations 4.3...
TRANSCRIPT
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4.1 Vector Spaces and Subspaces4.2 Null Spaces, Column Spaces, and Linear Transformations4.3 Linearly Independent Sets; Bases4.4 Coordinate systems
4 Vector Spaces
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Definition
Let H be a subspace of a vector space V. An indexed set of vectors in V is a basis for H if
i) is a linearly independent set, andii) the subspace spanned by coincides with H; i.e.
pbb ,,1
pbbH ,,Span 1
REVIEW
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The Spanning Set Theorem
Let be a set in V, and let .
a. If one of the vectors in S, say , is a linear combination of the remaining vectors in S, then the set formed from S by removing still spans H.
b. If , some subset of S is a basis for H.
pvvS ,,1 pvvH ,,Span 1
kv
kv
0H
REVIEW
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Theorem
The pivot columns of a matrix A form a basis for Col A.
REVIEW
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4.4 Coordinate Systems
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Why is it useful to specify a basis for a vector space?
One reason is that it imposes a “coordinate system” on the vector space.
In this section we’ll see that if the basis contains n vectors, then the coordinate system will make the vector space act like Rn.
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Theorem: Unique Representation Theorem
Suppose is a basis for V and is in V. Then there exists a unique set of scalars such that
.
pbb ,,1 pcc ,,1
ppcc bbx 11
x
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Definition:
Suppose is a basis for V and is in V. Thecoordinates of relative to the basis (the - coordinates of )are the weights such that .
pbb ,,1
pcc ,,1
If are the - coordinates of , then the vector in
is the coordinate vector of relative to , or the - coordinatevector of .
pcc ,,1
pc
c
x 1
xx x
x nR
xx
ppcc bbx 11
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Example:1. Consider a basis for , where
Find an x in such that .
2. For , find where is the standard basis for .
21,bb 2R
1
0,
1
221 bb
2R
3
2x
1
4x x 2R
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on standard basis on
1
4x
3
2x
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http://webspace.ship.edu/msrenault/ggb/linear_transformations_points.html
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The Coordinate Mapping
Theorem
Let be a basis for a vector space V. Then the coordinate mapping is a one-to-one and onto linear transformation from V onto .
€
x[ ]βx
€
[ ]
€
b1,L ,bn{ }
][xx nR
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1
4x
€
x[ ]β
1
2,
1
121 bb
Example:
For and , find .
For , let .
Then is equivalent to .
€
b1,L ,bp{ }
€
Pβ = b1,L ,bp[ ]
ppbcbcx 11
€
x = Pβ x[ ]β
: the change-of-coordinates matrix from β to the standard basis
P
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Example:
Let
Determine if x is in H, and if it is, find the coordinate vector of x relative to .
},&,{,
12
3
7
,
0
1
1
,
6
3
2
2121 vvxvv
}.,Span{ 21 vvH
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Application to Discrete MathLet = {C(t,0), C(t,1), C(t,2)} be a basis for
P2, so we can write each of the standard basis elements as follows:
C(t,0) = 1(1) + 0t + 0t2
C(t,1) = 0(1) + 1t + 0t2
C(t,2) = 0(1) – ½ t + ½ t2
This means that following matrix converts polynomials in the “combinatorics basis” into polynomials in the standard basis:
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Mβ =
1 0 0
0 1 −1/2
0 0 1/2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
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Application to Discrete MathRecall that = {C(t,0), C(t,1),
C(t,2)} is a basis for P2.
The following matrix converts polynomials in the “combinatorics basis” to polynomials in the standard basis:
Therefore, the following matrix converts polynomials in the the standard basis to polynomials in “combinatorics basis”:
€
Mβ−1 =
1 0 0
0 1 1
0 0 2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
€
Mβ =
1 0 0
0 1 −1/2
0 0 1/2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
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The polynomial p(t) = 3 + 5t – 7t2 has the following coordinate vector in the standard basis S = {1, t, t2}:
S
t
7
5
3
)(p
We now want to find the coordinate vector of p(t) in the “combinatorics basis” = {C(t,0), C(t,1), C(t,2)}:
14
2
3
7
5
3
200
110
001
That is, p(t) = 3 C(t,0) – 2 C(t,1) – 14 C(t,2)
Application to Discrete Math
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Application to Discrete MathFind a formula in terms of n for the following
sum
Solution. Using the form we found on the previous slide
€
3+ 5k − 7k 2( )
k =0
n
∑
€
3+ 5k − 7k 2
k =0
n
∑ = 3C(k,0) − 2C(k,1) −14C(k,2)k =0
n
∑
= 3C(n +1,1) − 2C(n +1,2) −14C(n +1,3)
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Aside: Why are these true?
http://webspace.ship.edu/deensley/DiscreteMath/flash/ch5/sec5_3/hockey_stick.html
€
C(k,2)k =0
n
∑ = C(n +1,3)€
C(k,0)k =0
n
∑ = C(n +1,1)
€
C(k,1)k =0
n
∑ = C(n +1,2)
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Application to Discrete MathFind the matrix M that
converts polynomials in the “combinatorics basis” into polynomials in the standard basis for P3:
Use this matrix to find a version of the following expression in terms of the standard basis:
€
3C(n +1,1) − 2C(n +1,2) −14C(n +1,3)
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Application to Discrete MathFind a formula in terms of n for the following
sum
Solution. Continued…..
€
3+ 5k − 7k 2( )
k =0
n
∑
€
3+ 5k − 7k 2
k =0
n
∑ = 3C(k,0) − 2C(k,1) −14C(k,2)k =0
n
∑
= 3C(n +1,1) − 2C(n +1,2) −14C(n +1,3)
= ___(n +1) + ___(n +1)2 + ___(n +1)3
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Final Steps…We can finish by multiplying by a matrix
that converts vectors written in {1,(t+1),(t+1)2,(t+1)3} coordinates to a vector written in terms of the standard basis.
Find such a matrix and multiply it by the answer on the previous slide to get a final answer of the form
€
___(1) + ___(n) + ___(n2) + ___(n3)