Page 311 - 35Linear Algebra
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17.2 Singular Value Decomposition                                                             311


                   are eigenvectors of
                                                          1   0 −  1
                                                                   
                                                          2        2
                                                  T
                                              MM =       0   2    0 
                                                        −  1  0    1
                                                          2        2
                   with eigenvalues 1 and 2, respectively. The third eigenvector (with eigenvalue 0) of
                   MM  T  is
                                                          1  
                                                          √
                                                            2
                                                   v 3 =  0 .
                                                             
                                                        
                                                          √ 1
                                                            2
                   The eigenvectors Mu 1 and Mu 2 are necessarily orthogonal, dividing them by their
                   lengths we obtain the left singular vectors and in turn our orthonormal output basis

                                                 1             1  
                                                   √        0       √
                                                    2                2
                                          0                     
                                                   0  ,  −1  ,  0 .
                                        O = 
                                                 − √ 1      0       √ 1
                                                    2                2
                                    0
                   The new matrix M of the linear transformation given by M with respect to the bases
                           0
                   O and O is
                                                              
                                                         1  √ 0
                                                    0
                                                 M =    0    2   ,
                                                         0    0
                                            √
                   so the singular values are 1, 2.
                                                                             0
                      Finally note that arranging the column vectors of O and O into change of basis
                   matrices
                                                                 1         1  
                                                                  √     0  √
                                          1     1  !               2         2
                                         √     √
                                          2      2                           
                                  P =               ,     Q =    0   −1    0   ,
                                         √ 1  − √ 1                          
                                          2      2                1         1
                                                                − √     0  √
                                                                   2         2
                   we have, as usual,
                                                     0
                                                  M = Q   −1 MP .

                      Singular vectors and values have a very nice geometric interpretation;
                   they provide an orthonormal bases for the domain and range of L and give
                   the factors by which L stretches the orthonormal input basis vectors. This
                   is depicted below for the example we just computed.


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