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10.2 Solution of X = AX for Constant A  307

                                        Linear combination of these solutions are also solutions. Thus, define two solutions
                                                                       1
                                                                        (  1 (t) +   2 (t))
                                                                       2
                                        and
                                                                       1
                                                                         (  1 (t) −   2 (t)),
                                                                       2i
                                        and these are the solutions given in the theorem.

                                           Theorem 10.7 enables us to replace two complex solutions

                                                                e (α+iβ) (U + iV) and e (α−iβ) (U − iV)
                                        in the general solution with the two solutions given in the theorem, which involve only real
                                        quantities.



                                 EXAMPLE 10.9
                                        We will solve the system X = AX with

                                                                         ⎛           ⎞
                                                                          2   0    1
                                                                     A = 0   −2   −2 ⎠ .
                                                                         ⎝
                                                                          0   2    0
                                                               √        √
                                        The eigenvalues are 2,−1 +  3i,−1 −  3i. Corresponding eigenvectors are, respectively,
                                                              ⎛ ⎞ ⎛          ⎞     ⎛         ⎞
                                                               1        1               1
                                                                        √              √
                                                                  ,            and  ⎝ 2 3i    .
                                                              ⎝ 0 ⎠ ⎝ −2 3i ⎠                ⎠
                                                                         √               √
                                                               0     −3 +  3i       −3 −   3i
                                        One solution is
                                                                           ⎛ ⎞
                                                                            1
                                                                                2t
                                                                               e .
                                                                           ⎝ 0 ⎠
                                                                            0
                                        Two other solutions are complex:
                                                         ⎛         ⎞            ⎛         ⎞
                                                              1                      1
                                                              √          √          √          √
                                                                     e (−1+ 3i)t  and  ⎝ 2 3i  e (−1− 3i)t .
                                                         ⎝ −2 3i ⎠                        ⎠
                                                                √                     √
                                                           −3 +  3i               −3 −  3i
                                        These three solutions can be used as columns of a fundamental matrix. However, if we wish, we
                                        can write a solution involving only real numbers and real-valued functions. First,
                                                                1          1         0
                                                           ⎛        ⎞   ⎛   ⎞    ⎛      ⎞
                                                                √                    √
                                                                      =       + i         = U + iV
                                                           ⎝ −2 3i ⎠    ⎝ 0 ⎠    ⎝ −2 3 ⎠
                                                                 √                  √
                                                            −3 +   3i     −3          3
                                        with
                                                                     ⎛   ⎞        ⎛      ⎞
                                                                       1              0
                                                                                      √
                                                                 U =  ⎝ 0 ⎠  and V = −2 3 ⎠ .
                                                                                  ⎝
                                                                                     √
                                                                      −3               3
                                                                        √
                                        By Theorem 10.7, with α =−1 and β =  3, we can replace the two complex solutions with the
                                        two solutions
                                                                         √          √
                                                                   −t
                                                                  e [cos( 3t)U − sin( 3t)V]
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                                   October 14, 2010  20:32  THM/NEIL   Page-307        27410_10_ch10_p295-342
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