Coordinate Systems
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A basis for a vector space V is an independent set of vectors that span V.
For example, in ℝ2 the standard basis is
1 0 
 '   
0 1 
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A basis for a vector space V is an independent set of vectors that span V.
For example, in ℝ2 the standard basis is
1 0 
 '   
0 1 
Suppose we have a different set of independent vectors from ℝ2.
For example, consider the set
1  1  
   '   
1  1 
Is this an alternate basis for ℝ2?
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A basis for a vector space V is an independent set of vectors that span V.
For example, in ℝ2 the standard basis is
1 0 
 '   
0 1 
Suppose we have a different set of independent vectors from ℝ2.
For example, consider the set
1  1  
   '   
1  1 
Is this an alternate basis for ℝ2?
The answer is YES. In fact any set of n independent vectors form a basis for ℝn.
Our task is to be able to switch back and forth between different bases for the
same vector space.
An example should help clarify the situation:
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Example:

8
2
Consider the vector x   
In the standard basis for ℝ2, our vector can be written as:
 8
1
0
x     8  2 
2
0
1
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Example:

8
2
Consider the vector x   
In the standard basis for ℝ2, our vector can be written as:
 8
1
0
x     8  2 
2
0
1
We would like to be able to write this vector in terms of our alternate basis
1  1  
   '   
1  1 
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Example:

8
2
Consider the vector x   
In the standard basis for ℝ2, our vector can be written as:
 8
1
0
x     8  2 
2
0
1
We would like to be able to write this vector in terms of our alternate basis
1  1  
   '   
1  1 
An equation would look something like this:
 8
1
1
x     c1     c2   
2
1
 1
Of course this type of equation should look familiar. It is just a 2x2 linear
system. We have solved this type of equation many times.
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 
Ac  x
1 1   c1   8
A
; c  c ; x   
1

1


2
 2
We can solve this by finding the inverse of the coefficient matrix.
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 
Ac  x
1 1   c1   8
A
; c  c ; x   
1

1


2
 2
We can solve this by finding the inverse of the coefficient matrix.
 




Ac  x  A 1  Ac  A 1  x  c  A 1  x
A 1 
1  1  1
 2  1 1 
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 
Ac  x
1 1   c1   8
A
; c  c ; x   
1

1


2
 2
We can solve this by finding the inverse of the coefficient matrix.
 




Ac  x  A 1  Ac  A 1  x  c  A 1  x
1  1  1
 2  1 1 

1  1  1 8 5
c
    


 2  1 1  2 3
A 1 
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 
Ac  x
1 1   c1   8
A
; c  c ; x   
1

1


2
 2
We can solve this by finding the inverse of the coefficient matrix.
 




Ac  x  A 1  Ac  A 1  x  c  A 1  x
1  1  1
 2  1 1 

1  1  1 8 5
c
    


 2  1 1  2 3
A 1 
So we have found our vector as a linear combination of the new basis vectors:
8
1
1

5


3

 
 
 
2
1
 1
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You may have noticed that the matrix we used was comprised of the basis
vectors in our alternate basis. Here is the notation they use in the textbook.


P x   x
1 1 
P  

1  1
In order to convert from the standard basis to the alternate basis, we found the
inverse of PB.
x 


 PB1 x
Vector written
in basis B
Vector written in
standard basis
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