Change of Basis


Consider the vector v  =  (2,5,3) in R3.  In writing these coordinates we mean 

        v  =  2e1 + 5e2 + 3e3 


        e1 =  (1,0,0)        e2 =  (0,1,0)        e3 =  (0,0,1)

are the standard basis vectors.  Sometimes we are interested in finding the coordinates with respect to another basis.  



Let S  =  {v1, v2, ... , vn} be a basis for a vector space V and let v be a vector in V and let

        v  =  c1v1 + ... + cnvn  

Then the coordinates of v with respect to the basis S is given by

        [v]S  =  (c1, ... , cn)



Consider the basis 

        S  =  {(1,2), (4,7)} 

of R2 and let 

        =  (5,8)

presented in the standard basis.  Find the coordinates of v in the basis S, that is find [v]S.  



We set

     (5,8)  =  c1(1,2) + c2(4,7)   


         c1 + 4c2  =  5
        2c1 + 7c2  =  8

We get the matrix equation


Notice that the matrix is just the matrix whose columns are the basis vectors of S.  The solution to this is



        c1  =  -3        c2  =  2

What we have seen here generalizes



Let S  =  {v1, v2, ... , vn} be a basis for a vector space V and let v be a vector in V.  Then

        [v]S  =  A-1v 

where A is the matrix whose column vectors are {v1, v2, ... , vn} 

The proof involves going over the previous example and generalizing.


It is also interesting to run this process in reverse.  Let  S  =  {v1, v2, ... , vn}be a basis for V and let [v]S be given.  We ask how to present v in the standard basis.  This follows from the theorem.  Since

        [v]S  =  A-1v 

We have

        v  =  A[v]S 




        S  =  {(1,3,4), (2,-1,1), (1,0,2)} 

be a basis for R3 and let 

        [v]S  =  (2,3,-1)

Find the coordinates with respect to the standard basis.  



We just find


so that 

        v  =  (7,3,9)


Let S  =  {(2,3), (1,4)} and T  =  {(0,2), (-1,5)} be two bases for R2, and let 

        [v]S  =  (-2,6)

Find [v]T 



We can first find v in the standard basis.  We have

        v  =  AS[v]S 

where AS is the matrix whose columns are the vectors in S.  Now convert to the T basis. 

        [v]T  =  (AT)-1v   =  (AT)-1AS[v]S    




Consider the vector v  =  2 + 3t - t2 and let S  =  {t, t - 1, t2 - 1}.  Find [v]S.  



This problem looks a lot different from the previous ones, but looks can be deceiving.  We notice that 

        t       =      0(1) + 1(t) + 0(t2)
        t - 1  =      -1(1) + 1(t) + 0(t2)
        t2 - 1  =      -1(1) + 0(t) + 1(t2)

We write that 


and use 


Transition Matrices

We have seen how to use the coordinates from one basis S into coordinates from another basis T.  We have

                [v]T  =  (AT)-1AS[v]S  

The matrix given by

        PT <-- S  =  (AT)-1AS 

is called the transition matrix from the S basis to the T basis.  Note that the transition matrix from the T basis to the S basis is given by 

        PS <-- T  =  (AS)-1AT  =  P-1T <-- S 



Find the transition matrix 

        PS <-- T  

for the bases of M2x2 given by


Then use this matrix to find
[v]S if 

        [v]T  =  (1,3,-2,4)


First we denote the standard basis by 

        E={1000,0100,0010,0001} read a11 a12 a21 a22

The AS is just the matrix of column vectors where each column is read as you would read the matrices in S.  That is


and similarly we have


The transition matrix is 


Now to find the coordinate in the S basis given T basis coordinates

        [v]T  =  (1,3,-2,4)

we just multiply


Remark:  The transition matrix will always be nonsingular because of the nonsingular equivalence and that S and T are linearly independent.

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