{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern