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4 Vector Spaces
4.4
COORDINATE SYSTEMS
© 2012 Pearson Education, Inc.
THE UNIQUE REPRESENTATION THEOREM
Theorem 7: Let B {b1 ,...,b n } be a basis for
vector space V. Then for each x in V, there exists a
unique set of scalars c1, …, cn such that
----(1)
x c1b1 ... cn bn
Proof: Since B spans V, there exist scalars such that
(1) holds.
Suppose x also has the representation
x d1b1 ... d n b n
for scalars d1, …, dn.
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Slide 4.4- 2
THE UNIQUE REPRESENTATION THEOREM
Then, subtracting, we have
0 x x (c1 d1 )b1 ... (cn d n )b n ----(2)
Since B is linearly independent, the weights in (2)
must all be zero. That is, c j d j for 1 j n .
Definition: Suppose B {b1 ,...,b n } is a basis for V
and x is in V. The coordinates of x relative to the
basis B (or the B-coordinate of x) are the weights
c1, …, cn such that x c1b1 ... cn b n .
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Slide 4.4- 3
THE UNIQUE REPRESENTATION THEOREM
If c1, …, cn are the B-coordinates of x, then the vector
n
c1
in
[x]B
cn
is the coordinate vector of x (relative to B), or the
B-coordinate vector of x.
The mapping x x B is the coordinate mapping
(determined by B).
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Slide 4.4- 4
COORDINATES IN
n
When a basis B for
is fixed, the B-coordinate
vector of a specified x is easily found, as in the
example below.
2
1
4
Example 1: Let b1 , b 2 , x , and
n
1
1
5
B {b1 , b 2 }. Find the coordinate vector [x]B of x
relative to B.
Solution: The B-coordinate c1, c2 of x satisfy
2
1 4
c1 c2
1
1 5
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b1
b2
x
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COORDINATES IN
or
n
2 1 c1 4
1 1 c 5
2
b1
b2
----(3)
x
This equation can be solved by row operations on an
augmented matrix or by using the inverse of the
matrix on the left.
In any case, the solution is c1 3 , c2 2 .
Thus x 3b1 2b 2 and
c1 3 .
x
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B
c2 2
Slide 4.4- 6
COORDINATES IN
n
See the following figure.
The matrix in (3) changes the B-coordinates of a
vector x into the standard coordinates for x.
An analogous change of coordinates can be carried
n
out in
for a basis B {b1 ,...,b n }.
Let PB
b1 b2
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bn
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COORDINATES IN
n
Then the vector equation
is equivalent to
x c1b1 c2 b2 ... cn b n
x PB x B
----(4)
PB is called the change-of-coordinates
matrix
n
from B to the standard basis in
.
Left-multiplication by PB transforms the coordinate
vector [x]B into x.
Since the columns of PB form a basis for
, PB is
invertible (by the Invertible Matrix Theorem).
n
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Slide 4.4- 8
COORDINATES IN
n
1
B
Left-multiplication by P converts x into its Bcoordinate vector:
P x x B
1
B
The correspondence x
1
x
B, produced by PB , is
the coordinate mapping.
1
Since PB is an invertible matrix, the coordinate
mapping is a one-to-one linear transformation from
n
onto
, by the Invertible Matrix Theorem.
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n
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THE COORDINATE MAPPING
Theorem 8: Let B {b1 ,...,b n } be a basis for a vector
x nB is a
space V. Then the coordinate mapping x
one-to-one linear transformation from V onto .
Proof: Take two typical vectors in V, say,
u c1b1 ... cn b n
w d1b1 ... d n b n
Then, using vector operations,
u v (c1 d1 )b1 ... (cn d n )b n
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Slide 4.4- 10
THE COORDINATE MAPPING
It follows that
u w
B
c1 d1 c1 d1
u B w B
cn d n cn d n
So the coordinate mapping preserves addition.
If r is any scalar, then
ru r (c1b1 ... cn b n ) (rc1 )b1 ... (rcn )b n
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THE COORDINATE MAPPING
So
ru
B
rc1
c1
r r u B
rcn
cn
Thus the coordinate mapping also preserves scalar
multiplication and hence is a linear transformation.
The linearity of the coordinate mapping extends to
linear combinations.
If u1, …, up are in V and if c1, …, cp are scalars, then
c1u1 ... c p u p c1 u1 B ... c p u p ----(5)
B
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B
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THE COORDINATE MAPPING
In words, (5) says that the B-coordinate vector of a
linear combination of u1, …, up is the same linear
combination of their coordinate vectors.
The coordinate mapping in Theorem 8 is an important
n
example of an isomorphism from V onto .
In general, a one-to-one linear transformation from a
vector space V onto a vector space W is called an
isomorphism from V onto W.
The notation and terminology for V and W may differ,
but the two spaces are indistinguishable as vector
spaces.
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Slide 4.4- 13
THE COORDINATE MAPPING
Every vector space calculation in V is accurately
reproduced in W, and vice versa.
In particular, any real vector space with a basis of n
vectors is indistinguishable from n .
3
1
3
Example 2: Let v1 6 , v 2 0 , x 12 ,
2
1
7
and B {v1 , v 2 }. Then B is a basis for
H Span{v1 ,v 2 } . Determine if x is in H, and if it is,
find the coordinate vector of x relative to B.
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Slide 4.4- 14
THE COORDINATE MAPPING
Solution: If x is in H, then the following vector
equation is consistent:
3
1 3
c1 6 c2 0 12
2
1 7
The scalars c1 and c2, if they exist, are the Bcoordinates of x.
© 2012 Pearson Education, Inc.
Slide 4.4- 15
THE COORDINATE MAPPING
Using row operations, we obtain
3 1 3
6 0 12
2 1 7
1 0 2
0 1 3 .
0 0 0
2
Thus c1 2 , c2 3 and [x]B .
3
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Slide 4.4- 16
THE COORDINATE MAPPING
The coordinate system on H determined by B is
shown in the following figure.
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Slide 4.4- 17