Page 231 - 35Linear Algebra
P. 231

12.1 Invariant Directions                                                                     231



                                               Reading homework: problem 2



                      Now that we’ve seen what eigenvalues and eigenvectors are, there are a
                   number of questions that need to be answered.



                      • How do we find eigenvectors and their eigenvalues?

                      • How many eigenvalues and (independent) eigenvectors does a given
                         linear transformation have?

                      • When can a linear transformation be diagonalized?


                   We will start by trying to find the eigenvectors for a linear transformation.




                                                 2 × 2 Example



                                         2
                                               2
                   Example 127 Let L: R → R such that L(x, y) = (2x + 2y, 16x + 6y). First, we
                   find the matrix of L, this is quickest in the standard basis:

                                                x   L     2 2     x
                                                   7−→               .
                                                y        16 6     y

                                                            x
                   We want to find an invariant direction v =    such that
                                                            y
                                                      Lv = λv

                   or, in matrix notation,


                                                  2 2     x           x
                                                               = λ
                                                 16 6     y           y

                                                  2 2     x         λ 0     x
                                    ⇔                          =
                                                 16 6     y         0 λ     y

                                          2 − λ    2      x         0
                                    ⇔                          =       .
                                            16   6 − λ    y         0
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