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282    CHAPTER 9  Eigenvalues, Diagonalization, and Special Matrices

                                    Thus far we have proved that, if A has n linearly independent eigenvectors, then A is diag-
                                              −1
                                 onalizable and P AP is the diagonal matrix having the eigenvalues down the main diagonal, in
                                 the order in which the eigenvectors are seen as columns of P.
                                    To prove the converse, now suppose that A is diagonalizable. We want to show that A has
                                 n linearly independent eigenvectors (regardless of whether the eigenvalues are distinct). Further,
                                                      −1
                                 we want to show that, if Q AQ is a diagonal matrix, then the diagonal elements of this matrix
                                 are the eigenvalues of A, and the columns of Q are corresponding eigenvectors. Thus suppose
                                 that
                                                                      0       0
                                                                 ⎛              ⎞
                                                                  d 1     ···
                                                                   0      ···  0
                                                                 ⎜    d 1       ⎟
                                                          −1     ⎜              ⎟
                                                        Q AQ = ⎜ .    .   .   . ⎟ = D.
                                                                 ⎝ . .  . .  . .  . . ⎠
                                                                   0  0   ··· d n
                                 Let V j be column j of Q. These columns are linearly independent because Q is nonsingular. We
                                 will show that d j is an eigenvalue of A with eigenvector V j .
                                          −1
                                    From Q AQ=D,wehave AQ=QD. Compute both sides of this equation separately. First,

                                 since the columns of Q are the V j s, then
                                                                          ⎛              ⎞
                                                                                0  ···  0
                                                                        ⎞ d 1
                                                        ⎛
                                                          |    |  ···  |
                                                                          ⎜  0  d 1  ···  0  ⎟
                                                                                         ⎟
                                                                          ⎜
                                                   QD = V 1   V 2  ··· V n  ⎠ ⎜ .  .  .  . ⎟
                                                        ⎝
                                                                          ⎝ .   .   .
                                                          |    |  ···  |    .   .   .   . . ⎠
                                                                            0   0  ··· d n
                                                        ⎛                     ⎞
                                                           |     |   ···   |
                                                                               ,
                                                      = d 1 V 1  d 2 V 2  ··· d n V n  ⎠
                                                        ⎝
                                                           |     |   ···   |
                                 which is a matrix having d j V j as column j. Now compute
                                                     ⎛               ⎞   ⎛                   ⎞
                                                       |    |  ···  |       |    |   ···   |
                                                                                               ,
                                              AQ = A V 1   V 2  ··· V n  ⎠  = AV 1  AV 2  ··· AV n  ⎠
                                                                         ⎝
                                                     ⎝
                                                       |    |  ···  |       |    |   ···   |
                                 which is a matrix having AV j as column j. Since these matrices are equal, then
                                                                  AV j = d j V j
                                 and this makes d j an eigenvalue of A with eigenvector V j .
                                    Not every matrix is diagonalizable. We know from the theorem that a n × n matrix with
                                 fewer than n linearly independent eigenvectors is not diagonalizable.
                         EXAMPLE 9.11
                                 Let

                                                                     1  −1
                                                                 B =         .
                                                                     0   1
                                 B has eigenvalues 1,1, and all eigenvectors are constant multiples of

                                                                      1
                                                                         .
                                                                      0
                                 Therefore B has as eigenvectors only nonzero multiples of one vector, and does not have two
                                 linearly independent eigenvectors. By the theorem, B is not diagonalizable.




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                                   October 14, 2010  14:49  THM/NEIL   Page-282        27410_09_ch09_p267-294
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