Page 327 - 35Linear Algebra
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327


                                                        T
                                 T
                         Since M M  −1  6= I, it follows M 6= M so M is not symmetric. Finally

                                                                     2   1    2    1
                                                        2
                                     T
                              trf(M) = trf(M) = tr(M − I) = tr                        − trI
                                                                     3 −1     3 −1
                                      = (2 · 2 + 1 · 3) + (3 · 1 + (−1) · (−1)) − 2 = 9 .
                      5. First

                                                                     cos θ  sin θ  x
                                                  T

                                   X (MX) = X MX = x y
                                                                   − sin θ cos θ   y

                                                        x cos θ + y sin θ
                                                                            2   2
                                           = x y                        = (x + y ) cos θ .
                                                      −x sin θ + y cos θ
                                     √         p
                                                                                  T
                                                       2
                                                   2
                         Now ||X|| =   X X =     x + y and (MX) (MX) = XM MX. But

                                                   cos θ − sin θ     cos θ  sin θ
                                           T
                                        M M =
                                                   sin θ   cos θ   − sin θ cos θ
                                               2       2
                                             cos θ + sin θ              0
                                         =                     2       2   = I .
                                                         0 cos θ + sin θ
                                                p
                                                        2
                                                   2
                         Hence ||MX|| = ||X|| =   x + y . Thus the cosine of the angle between X
                         and MX is given by
                                                           2
                                                                2
                                         X (MX)          (x + y ) cos θ
                                                                          = cos θ .
                                                    = p         p
                                       ||X|| ||MX||      x + y 2  x + y 2
                                                          2
                                                                    2
                         In other words, the angle is θ OR −θ. You should draw two pictures, one
                         where the angle between X and MX is θ, the other where it is −θ.
                                                |X (MX)|
                         For Cauchy–Schwartz,             = | cos θ| = 1 when θ = 0, π. For the
                                               ||X|| ||MX||
                         triangle equality MX = X achieves ||X + MX|| = ||X|| + ||MX||, which
                         requires θ = 0.
                      6. This is a block matrix problem. Notice the that matrix M is really just

                                I I
                         M =         , where I and 0 are the 3×3 identity zero matrices, respectively.
                                0 I
                         But

                                                    I I     I I       I 2I
                                              2
                                            M =                   =
                                                    0 I     0 I       0   I
                         and

                                                   I I     I 2I        I 3I
                                              3
                                            M =                    =
                                                   0 I     0   I       0   I
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