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Multiplicity and formulas for the trace and determinant             109



                  unit length if desired. Consider the example
                                21      D = det(A) = (2)(3) − (1)(1) = 6 − 1 = 5

                           A =                                                       (3.34)
                                13                T = tr(A) = 2 + 3 = 5
                  with the eigenvalues
                                                                          √
                              1     1            1     1              5    5
                                        2
                        λ 1,2 =  T ±  T − 4D =    (5) ±   25 − (4)(5) =  ±           (3.35)
                              2     2            2     2              2    2
                  An eigenvector w [1]  for
                                                         √
                                                     5    5
                                                λ 1 =  +
                                                     2    2
                  must satisfy
                                                     )     √ *
                                                       1    5
                                   [1]     [1]                   [1]   [1]
                         (a 11 − λ 1 )w  1  + a 12 w  2  = 0 ⇒−  +  w 1  + w  2  = 0  (3.36)
                                                       2    2
                           [1]               [1]
                  If we set w  = 1, and compute w  from (3.36), we obtain
                           1                 2
                                                           √ *
                                                 
   )          T
                                                       1    5
                                            [1]
                                           w  = 1       +                            (3.37)
                                                       2   2
                  Above, we have selected a unique eigenvector by arbitrarily setting the value of one com-
                  ponent. More generally, we may select some vector v, say at random, and choose as a
                  normalization, v · w [ j]  = 1. We replace row m of the matrix (A − λ j I) with the elements
                  of v to obtain a modified matrix (A − λ j I) ++ . We then obtain the eigenvector by solving
                  the linear system
                                      (A − λ j I) ++ w [ j]  = e [m]  e [m]          (3.38)
                                                               j  = δ mj
                  Shortly, we consider numerical methods to compute λ k and w  [k]  without the need for
                  tedious analytical calculations.



                  Multiplicity and formulas for the trace and determinant


                  For an N × N matrix A, the characteristic polynomial
                                         N  N     N

                     p(λ) = det(A − λI) =     ...    ε i 1 ,i 2 ,...,i N  (a i 1 ,1 − λδ i 1 ,1 ) ... (a i N ,N − λδ i N ,N )
                                        i 1 =1 i 2 =1  i N =1
                                                                                     (3.39)
                  is of degree N and thus has N roots, so an N × N matrix has N eigenvalues. However, these
                  may not all be distinct; i.e., we may have λ j = λ k for some j  = k. As an example, consider
                  the matrix
                                                
                                        2   3  4                  2
                                                      p(λ) = (2 − λ) (1 − λ)
                                  A =   0  2  5                                    (3.40)
                                                     λ 1 = 2  λ 2 = 2  λ 3 = 1
                                        0   0  1
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