Page 337 - 35Linear Algebra
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                         The associated eigenvalues solve the homogeneous systems (in augmented
                         matrix form)

                            1 −5 0           1 −5 0              5 −5 0            1 −1 0
                                         ∼                and                 ∼                ,
                            1 −5 0           0   0   0           1 −1 0            0  0   0

                                                   5               1
                                                                                        12
                                                                                12
                         respectively, so are v 2 =   and v −2 =     . Hence M v 2 = 2 v 2 and
                                                   1               1

                                                     x     x−y   5    x−5y  1
                           12
                                        12
                         M v −2 = (−2) v −2 . Now,       =          −          (this was obtained
                                                     y      4    1     4    1
                         by solving the linear system av 2 + bv −2 = for a and b). Thus

                                               x     x − y        x − 5y
                                           M       =       Mv 2 −       Mv −2
                                                y      4             4

                                               x − y     x − 5y           x
                                         = 2 12      v 2 −      v −2 = 2 12     .
                                                 4          4               y
                         Thus

                                                          4096      0
                                                    12
                                                 M    =                .
                                                             0 4096
                         If you understand the above explanation, then you have a good understanding

                                                                                          4 0
                                                                                     2
                         of diagonalization. A quicker route is simply to observe that M =      .
                                                                                          0 4
                      8.

                                                    a − λ     b
                                               2
                                 P M (λ) = (−1) det                = (λ − a)(λ − d) − bc .
                                                       c    d − λ
                         Thus
                                            P M (M) = (M − aI)(M − dI) − bcI

                                      a  b      a 0        a   b     d 0         bc   0
                                 =          −                     −          −
                                      c d       0 a         c d       0 d         0 bc

                                            0    b     a − d b       bc   0
                                        =                        −           = 0 .
                                            c d − a      c    0       0 bc
                         Observe that any 2 × 2 matrix is a zero of its own characteristic polynomial
                         (in fact this holds for square matrices of any size).
                                                                                   2
                                                     2
                         Now if A = P  −1 DP then A = P     −1 DPP −1 DP = P   −1 D P. Similarly
                                    k
                          k
                         A = P  −1 D P. So for any matrix polynomial we have
                                       n
                                      A + c 1 A n−1  + · · · c n−1 A + c n I
                                           n
                                  = P  −1 D P + c 1 P −1 D n−1 P + · · · c n−1 P −1 DP + c n P  −1 P
                                            n
                                  = P  −1 (D + c 1 D n−1  + · · · c n−1 D + c n I)P .
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