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5.6 Unitary and Orthogonal Matrices                                                335
                   Example 5.6.6
                                                      n
                                    Problem: If x ∈     is a vector such that  x  =1, explain how to use plane
                                                                               n
                                    rotations to construct an orthonormal basis for    that contains x.
                                    Solution: This is almost the same problem as that posed in Example 5.6.3, and,
                                    as explained there, the goal is to construct an orthogonal matrix Q such that
                                    Q ∗1 = x. But this time we need to use plane rotations rather than an elementary
                                    reflector. Equation (5.6.17) asserts that we can build an orthogonal matrix from
                                    a sequence of plane rotations P = P 1n ··· P 13 P 12 such that Px = e 1 . Thus
                                          T
                                                 T
                                    x = P e 1 = P , and hence the columns of Q = P T  serve the purpose. For
                                                 ∗1
                                    example, to extend
                                                                          
                                                                        −1
                                                                x =  1  2 
                                                                           
                                                                      
                                                                    3     0
                                                                        −2
                                                               4
                                    to an orthonormal basis for   , sequentially annihilate the second and fourth
                                    components of x by using (5.6.16) to construct the following plane rotations:
                                                         √       √                         √
                                                     −1/ 5     2/ 500           −1           5
                                                                                         
                                                         √       √
                                                    −2/ 5   −1/ 500  1  2          1    0 
                                                        0       0     1  0  3    0     3     0
                                            P 12 x =                             =        ,
                                                        0       0     0  1      −2         −2
                                                           √                       √
                                                             5/300      −2/3         5      1
                                                                                       
                                                             0    1  0    0
                                                                             1    0    0 
                                            P 14 P 12 x =                           =   .
                                                             0    0  1    0    3     0      0
                                                                        √
                                                            2/3   0 0     5/3      −2       0
                                    Therefore, the columns of
                                                                                 √           √
                                                                       −1/3  −2/ 50     −2/3 5
                                                                                               
                                                                                 √           √
                                                        T
                                                             T
                                           Q =(P 14 P 12 ) = P P T    2/3   −1/ 50       4/3 5 
                                                             12  14  =   0     0    1      0   
                                                                                          √
                                                                       −2/3     0    0      5/3
                                    are an orthonormal set containing the specified vector x.
                   Exercises for section 5.6
                                    5.6.1. Determine which of the following matrices are isometries.
                                                        √       √
                                                                                               
                                                      1/ 2   −1/ 2      0                10      1
                                                        √       √        √
                                               (a)    1/ 6   1/ 6   −2/ 6   .   (b)    10   −1    .
                                                        √       √        √
                                                      1/ 3    1/ 3     1/ 3              01      0
                                                                                               
                                                                            e iθ 1  0      0
                                                      0010                              ···
                                                                             0    e iθ 2  ···  0 
                                               (c)   1000            (d)    .    .   .    .   .
                                                                 .
                                                    
                                                      0001                   . .   . .  . .  . . 
                                                      0100                     0    0   ··· e iθ n
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