Change of Basis Coordinates Consider the vector v  =  (2,5,3) in R3.  In writing these coordinates we mean          v  =  2e1 + 5e2 + 3e3  Where          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.     Definition 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)   Example Consider the basis          S  =  {(1,2), (4,7)}  of R2 and let          v  =  (5,8) presented in the standard basis.  Find the coordinates of v in the basis S, that is find [v]S.     Solution We set      (5,8)  =  c1(1,2) + c2(4,7)    or          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 or         c1  =  -3        c2  =  2 What we have seen here generalizes   Theorem 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    Example Let          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.     Solution We just find so that          v  =  (7,3,9) Example 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    Solution 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     or Example Consider the vector v  =  2 + 3t - t2 and let S  =  {t, t - 1, t2 - 1}.  Find [v]S.     Solution 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    Example 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) Solution First we denote the standard basis by 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. Back to the Linear Algebra Home Page