5.3.  Coordinates and change of basis in a vector space

Let V be a  vector space over with a basis = {}. Then, for every v V , there exist unique scalars such that v = . Let us denote these scalars by . If we fix the order of elements of  , then for every v V we get a unique element () , where
                                                          v = .

Thus to every v V , we can associate an element () . This motivates our next condition.

5.3.1 Definition:

Let V be a vector space and = {} be an ordered basis of V, (i.e., is a basis and the order of its elements is fixed). For every v V, the unique scalars such that  v = are called the coordinate functions and =  () is called the coordinate vector will change if the ordered basis .
             Note that, the coordinate vector will change if the order of elements of  is changed.

 
5.3.2 Example:

(i)    Let = {} be the natural of . Then for any v >(x, y, z) , v = .
       Thus,
                       
       Similar statements holds for .

(ii)   Let V = , the vector space of all real polynomial of degree at most 3. Let = {},
      . is the standard basis for . It is early to check that is also basis for .
       For any vector p(x) = ,
                           
       Note that, = if and only if
                                        a =
                                        b = -
                                        c = 2 + +
      Hence, = a, = (), = (). Thus for p(x) = ,


                               

     An element v V uniquely determines and is determined by its coordinate vector. In fact, we have the
     following :

5.3.3 Theorem:

Let V be a vector space and = {} be an ordered basis of V. Then the map V from V to is a bijective map such that for

(i) 
(ii)

 Proof

 
5.3.4 Notes:

The corresponding  V is one-to-one and identifies the vector space V with the space in all respects. Thus all statements regarding any vector space of dimension n can be analyzed in the vector space in terms of the coordinate vectors once an ordered basis of V is chosen and fixed.

 
5.3.5 Theorem:

Vectors {} is linearly independent in V, if and only if the corresponding coordinate vectors is linearly independent in .

 Proof

 
5.3.6 Theorem:

Let V be a vector space over and be two ordered basis of V. For any v V , let be its coordinate vector with respect to basis and be its coordinate vector with respect to basis . For every i, 1 i n, let the coordinate vector , with respect to the basis be given by , and let . Then P is an invertible matrix and P = .

 Proof

 
5.3.7 Notes:

Above theorem has a converse: let  be any given invertible matrix and = be any given basis of a vector space V. Define
                            , 1 i n,
where  . Then it is easy to such that = {} is a basis of V and the change of coordinates from basis to is precisely given by (5.1), that is by the matrix P. Hence change of basis is equivalent to multiplication by an invertible matrices. Let us look at an example.

 
5.3.8 Example:

Consider the vector space V = over . Let = and := . Note that is the standard basis and it is easy to check that is also a basis of  . Since
                            
and
                            
the change of basis from and is given by the matrix
                                
Thus, for a point P having coordinates (x, y) with respect to , its coordinate with respect to new basis are :
                                                  
                   
                                                                           (Figure 5.1)

Similarly, the matrix corresponding to the change from to is given by the matrix


                                                  
Clearly,
                                 

 
5.3.9 Example:

Consider the standard basis = and consider the new basis given by the matrix i.e., the basis {u, v} where
                            
and
                            



                                                                              (Figure 5.2)


It is easy to check that {, } is indeed a basis of  . Geometrically, the new basis is obtained from the earlier by rotation through an angle .

 
5.3.10 Example:

Let V = , = {(2, 0, 1), (1, 2, 0), (1, 1, 1)} and = . We ask the reader to verify that both and are bases for . Let v = (4, -9, 5) . Let us first complete and . Let v = , i.e.,
                                        
It is early to see that this implies = 4, = -5, = 1. Similarly, if v = , i.e.,
                                       
then a = 1, b = 2, c = 3. Hence
                             
Next to find the matrix P such that
                              
We recall that the column of P is the coordinate vector if  with respect to the basis . This means we have to solve the equations
                                      
Note that each is a system of linear equations:
                          
Since then system has same coefficient matrix, we can show these together. Consider the augmented matrix to reduced row-echelon form. This (we leave the calculation to the reader) is
                     
Thus
                   
Using (5.2) and (5.4), it is easy to check that equation (5.3) holds.
              Similarly, the matrix Q that hold gives the change of conditions from to , i.e., , can be computed by solving.
                                 
to get
                                                 
Finally, we ask the reader to verify that PQ = QP = Id.
(iii) Let ={(1, -3), (-2, 4)} be a basis of  and let
                       
Let us find a basis of  such that the matrix for the change of basis from to is P. For this, we note that
                     
Hence, = {(-7, 9), (-5, 7)}.

 
Click here to see an interactive visualization: Applet 5.3
Click here to take a Quiz: Quiz 5.3