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5.9 Complementary Subspaces                                                        389
                   Example 5.9.2

                                    Angle between Complementary Subspaces.       The angle between nonzero
                                                         n
                                    vectors u and v in     was defined on p. 295 to be the number 0 ≤ θ ≤
                                                          T
                                    π/2 such that cos θ = v u/  v   u  . It’s natural to try to extend this idea
                                                                 2   2
                                                                                        n
                                    to somehow make sense of angles between subspaces of   . Angles between
                                    completely general subspaces are presently out of our reach—they are discussed
                                    in §5.15—but the angle between a pair of complementary subspaces is within
                                                     n
                                    our grasp. When   = R⊕N with R = 0  = N, the angle (also known as the
                                    minimal angle) between R and N is defined to be the number 0 <θ ≤ π/2
                                    that satisfies
                                                                    T
                                                                  v u                T
                                                     cos θ = max         =    max   v u.          (5.9.16)
                                                            u∈R  v   u  2   u∈R, v∈N
                                                                   2
                                                            v∈N             u  = v  =1
                                                                             2    2
                                    While this is a good definition,it’s not easy to use—especially if one wants to
                                    compute the numerical value of cos θ. The trick in making θ more accessible
                                                                                     2
                                    is to think in terms of projections and sin θ =(1 − cos θ) 1/2 . Let P be the
                                    projector such that R (P)= R and N (P)= N, and recall that the matrix
                                    2-norm (p. 281) of P is
                                                              P  = max  Px  .                     (5.9.17)
                                                                2             2
                                                                     x  =1
                                                                      2
                                    In other words,  P  is the length of a longest vector in the image of the unit
                                                      2
                                    sphere under transformation by P. To understand how sin θ is related to  P  ,
                                                                                                       2
                                                           3
                                    consider the situation in   . The image of the unit sphere under P is obtained
                                    by projecting the sphere onto R along lines parallel to N. As depicted in
                                    Figure 5.9.2,the result is an ellipse in R.



                                                                   P
                                                      v
                                                         =  max  Px   =
                                                          x =1
                                                                 x
                                                                  θ         θ
                                                            v








                                                                  Figure 5.9.2
                                    The norm of a longest vector v on this ellipse equals the norm of P. That is,
                                     v  = max  x  =1  Px  =  P  , and it is apparent from the right triangle in
                                       2         2       2      2
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