Page 359 - 35Linear Algebra
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359


                              Since this (spanning) set has three (linearly independent) elements,
                                                              3
                              the vector space of linear maps B → B has dimension 3. This is an
                              example of a general notion called the dual vector space.
                                                               
                                              a                   a
                                      2
                               2
                                                        2
                                2 =
                                                                  b
                                               b
                     11.  (a)  d X   d cos(ωt)     = −ω cos(ωt)     .
                              dt       dt 2
                                               c                  c
                              Hence
                                                                                      
                                                −a − b                     −1 −1      0    a
                                                                                           b
                                F = cos(ωt)    −a − 2b − c    = cos(ωt)   −1 −2 −1    
                                                 −b − c                     0 −1 −1        c
                                                                              
                                                                               a
                                                                     2
                                                              = −ω cos(ωt)       ,
                                                                               b
                                                                               c
                              so
                                                                      
                                                           −1 −1      0
                                                    M =   −1 −2 −1      .
                                                             0 −1 −1
                          (b)
                                                     
                                   λ + 1    1      0

                              det    1   λ + 2    1     = (λ + 1) (λ + 2)(λ + 1) − 1 − (λ + 1)
                                     0      1    λ + 1

                                                          = (λ + 1) (λ + 2)(λ + 1) − 2
                                                                       2
                                                          = (λ + 1) λ + 3λ) = λ(λ + 1)(λ + 3)
                              so the eigenvalues are λ = 0, −1, −3.
                              For the eigenvectors, when λ = 0 we study:
                                                                                     
                                             −1 −1       0      1    1    0       1 0 −1
                                 M − 0.I =   −1 −2 −1      ∼   0 −1 −1    ∼   0 1   1   ,
                                               0 −1 −1          0 −1 −1           0 0    0
                                    
                                    1
                              so   −1   is an eigenvector.
                                    1
                              For λ = −1
                                                                                
                                                          0 −1      0      1 0 1
                                        M − (−1).I =   −1 −1 −1       ∼   0 1 0   ,
                                                          0 −1      0      0 0 0


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