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4.8 Change of Basis and Similarity                                                 255


                                    and the matrix of A with respect to S is

                                                                2  −1      01      1  1      10
                                                  −1
                                         [A] S   = Q  [A] S Q =                          =         .
                                                              −1    1    −23       1  2      02
                                    Notice that [A] S   is a diagonal matrix, whereas [A] S is not. This shows that
                                    the standard basis is not always the best choice for providing a simple matrix
                                    representation. Finding a basis so that the associated coordinate matrix is as
                                    simple as possible is one of the fundamental issues of matrix theory. Given an
                                    operator A, the solution to the general problem of determining a basis B so
                                    that [A] B is diagonal is summarized on p. 520.

                   Example 4.8.3
                                                                                             n
                                    Problem: Consider a matrix M n×n to be a linear operator on    by defining
                                                                                                       n
                                    M(v)= Mv (matrix–vector multiplication). If S is the standard basis for   ,
                                    and if S = {q 1 , q 2 ,..., q n } is any other basis, describe [M] S and [M] S  .

                                    Solution: The j th  column in [M] S is [Me j ] S =[M ∗j ] S = M ∗j , and hence
                                    [M] S = M. That is, the coordinate matrix of M with respect to S is M itself.
                                    To find [M] S  , use (4.8.6) to write [M] S   = Q −1 [M] S Q = Q −1 MQ, where



                                           Q =[I] S   S =  [q 1 ] S [q 2 ] S ··· [q n ] S  = q 1 q 2 ··· q n .






                                    Conclusion: The matrices M and Q −1 MQ represent the same linear operator
                                    (namely, M), but with respect to two different bases (namely, S and S ). So,

                                    when considering properties of M (as a linear operator), it’s legitimate to replace
                                    M by Q  −1 MQ. Whenever the structure of M obscures its operator properties,
                                    look for a basis S = {Q ∗1 , Q ∗2 ,..., Q ∗n } (or, equivalently, a nonsingular matrix

                                    Q) such that Q −1 MQ has a simpler structure. This is an important theme
                                    throughout linear algebra and matrix theory.
                                        For a linear operator A, the special relationships between [A] B and [A] B
                                    that are given in (4.8.5) and (4.8.6) motivate the following definitions.

                                                                Similarity

                                       •   Matrices B n×n and C n×n are said to be similar matrices when-
                                           ever there exists a nonsingular matrix Q such that B = Q −1 CQ.
                                           We write B - C to denote that B and C are similar.
                                                                 n×n     n×n                    −1
                                       •   The linear operator f :    →      defined by f(C)= Q    CQ
                                           is called a similarity transformation.
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