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lecture_36 - MA 265 LECTURE NOTES FRIDAY APRIL 11...

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MA 265 LECTURE NOTES: FRIDAY, APRIL 11 Eigenvalues and Eigenvectors Review. Let ( V, + , · ) be a real vector space of dimension n . Recall that a linear operator is a function L : V V such that L preserves vector addition: For all u , v V , we have L ( u + v ) = L ( u ) + L ( v ). L preserves scalar multiplication: For all scalars c R and vectors u V , we have L ( c u ) = c L ( u ). In practice, we will choose V = R n . Recall that there is an n × n matrix A , the standard matrix representing L , such that L ( x ) = A x . We would like to find all vectors x V such that L ( x ) is parallel to x , i.e., L ( x ) = λ x for some scalar λ . Example. Consider V = R 2 . As in the last lecture, fix a positive scalar r and consider the linear operator L : R 2 R 2 defined as L ( x ) = A x in terms of the 2 × 2 matrix A = r 0 0 r = L x y = r x r y . Recall that this is either a dilation (if r > 1) or a contraction (if 0 < r < 1). Hence L ( x ) = λ x for all x R 2 where λ = r . Note that A is a scalar matrix. Example. As another example along similar lines, consider the 2 × 2 matrix A = 1 0 0 - 1 = L x y = x - y . Recall that this is the reflection about the x -axis. In particular, there is no scalar λ so that L ( x ) = λ x for all vectors x V . We will show, however, that there are subspaces W of V = R 2 where we can find such a scalar. Define W and W as those subspaces of V which are the x -axis and y -axis, respectively: W = x = x y y = 0 = span 1 0 W = x = x y x = 0 = span 0 1 It is easy to see that W is the orthogonal complement to W , so that we have the direct sum V = W
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