Page 361 - 35Linear Algebra
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                          (b) Yes, the matrix M is symmetric, so it will have a basis of eigenvectors
                              and is similar to a diagonal matrix of real eigenvalues.

                                                                     a − λ      b
                              To find the eigenvalues notice that det               = (a − λ)(d −
                                                                         b d − λ
                                                     2


                                    2
                              λ) − b = λ −   a+d 2  − b −  a−d 2 . So the eigenvalues are
                                              2            2
                                             r                               r
                                     a + d           a − d  2        a + d          a − d  2
                                                                                 2
                                                2
                                 λ =       +   b +           and µ =       −    b +          .
                                       2               2                2              2
                          (c) The trick is to write
                                T
                                         T
                                                          T
                                                               T
                                                 T
                                                                                        T
                              X MX +C X +X C = (X +C M            −1 )M(X +M   −1 C)−C M    −1 C ,
                              so that
                                           T
                                                 T
                                                                           T
                                        (X + C M     −1 )M(X + M  −1 C) = C MC − f .
                                                               T
                              Hence Y = X + M   −1 C and g = C MC − f.
                          (d) The cosine of the angle between vectors V and W is given by
                                                                      T
                                                    V W             V W
                                                √             = √             .
                                                                          T
                                                 V V W W          V V W W
                                                                    T
                                                                                             T
                              So replacing V → PV and W → PW will always give a factor P P
                                                          T
                              inside all the products, but P P = I for orthogonal matrices. Hence
                              none of the dot products in the above formula changes, so neither does
                              the angle between V and W.
                          (e) If we take the eigenvectors of M, normalize them (i.e. divide them
                              by their lengths), and put them in a matrix P (as columns) then P
                              will be an orthogonal matrix. (If it happens that λ = µ, then we
                              also need to make sure the eigenvectors spanning the two dimensional
                              eigenspace corresponding to λ are orthogonal.) Then, since M times
                              the eigenvectors yields just the eigenvectors back again multiplied by
                              their eigenvalues, it follows that MP = PD where D is the diagonal
                              matrix made from eigenvalues.
                                                             T
                                                  T
                                                                             T
                                                                                            T
                                                                T
                                                                                T
                          (f) If Y = PZ, then Y MY = Z P MPZ = Z P PDZ = Z DZ

                                          λ   0
                              where D =          .
                                           0 µ
                          (g) Using part (f) and (c) we have
                                                          2
                                                                 2
                                                        λz + µw = g .
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