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5.6 Unitary and Orthogonal Matrices                                                325




                                                 Properties of Elementary Reflectors

                                       •   All elementary reflectors R are unitary, hermitian, and involutory
                                             2
                                           ( R = I ). That is,
                                                               R = R = R   −1 .                 (5.6.9)
                                                                     ∗
                                       •   If x n×1 is a vector whose first entry is x 1  =0, and if

                                                                         1     if x 1 is real,
                                        u = x ± µ  x  e 1 ,  where  µ =                        (5.6.10)
                                                                       x 1 /|x 1 | if x 1 is not real,
                                           is used to build the elementary reflector R in (5.6.7), then

                                                               Rx = ∓µ  x  e 1 .               (5.6.11)

                                           In other words, this R “reflects” x onto the first coordinate axis.
                                           Computational Note: To avoid cancellation when using floating-
                                           point arithmetic for real matrices, set u = x + sign(x 1 )  x  e 1 .


                                                                         ∗
                                    Proof of (5.6.9). It is clear that R = R , and the fact that R = R −1  is
                                                                      2
                                    established simply by verifying that R = I.
                                    Proof of (5.6.10). Observe that R = I − 2ˆ uˆ u , where ˆ u = u/  u  .
                                                                           ∗
                                    Proof of (5.6.11). Write Rx = x − 2uu x/u u = x − (2u x/u u)u and verify
                                                                                           ∗
                                                                       ∗
                                                                           ∗
                                                                                       ∗
                                    that 2u x = u u to conclude Rx = x − u = ∓µ  x  e 1 .
                                                 ∗
                                           ∗
                   Example 5.6.3
                                                        n×1
                                    Problem: Given x ∈C      such that  x  =1, construct an orthonormal basis
                                        n
                                    for C  that contains x.
                                    Solution: An efficient solution is to build a unitary matrix that contains x as
                                                                              ∗
                                    its first column. Set u = x±µe 1 in R = I−2(uu /u u) and notice that (5.6.11)
                                                                                 ∗
                                    guarantees Rx = ∓µe 1 , so multiplication on the left by R (remembering that
                                     2
                                    R = I) produces x = ∓µRe 1 =[∓µR]       . Since |∓ µ| =1, U = ∓µR
                                                                          ∗1
                                    is a unitary matrix with U ∗1 = x, so the columns of U provide the desired
                                                                                                   4
                                    orthonormal basis. For example, to construct an orthonormal basis for   that
                                                                   T
                                    includes x =(1/3) ( −120 − 2) , set
                                                    −4                                   −120      −2
                                                                                                   
                                                                              uu T   1  220         1 
                                                1  2 
                                    u = x − e 1 =      and compute R = I − 2     =                  .
                                                                                T
                                                3    0                        u u    3    003        0
                                                    −2                                   −210        2
                                    The columns of R do the job.
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