MIT18_024S11_Pset2

MIT18_024S11_Pset2 - 4 1 − 2 A = 16 − 2 − 8 ⎞ ⎝ 4...

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

View Full Document Right Arrow Icon
PSET 2 - DUE FEBRUARY 15 1. Let T 2 2 θ : R R be the linear transformation that takes a vector x to its rotation (counterclockwise) by θ degrees about the origin. a. Find the matrix representation for T using the standard basis for R 2 θ ( { (1 , 0) , (0 , 1) } ). (4 pts) b. This matrix is obviously invertible. Find its inverse and verify by matrix multiplica- tion. (2 pts) 2.Let T : R 3 R 2 and S : R 3 R 3 correspond to the transformations T ( x, y, z ) = ( x, y ); S ( x, y, z ) = ( x, y, z ) . Notice that TS is a well defined linear transformation. Find a matrix representation for S , T , and TS using the basis (1 , 0 , 0) , (1 , 1 0) , , (1 , 1 , 1) for R 3 and the basis (1 , 0) , (1 , 1) for R 2 { } { } . (6 pts) 3. 2.20:9 (4 pts) 4. For an n × n matrix A , we define λ R to be an eigenvalue of A if Ax = λx for some x = 0 R n . a. Prove that λ is an eigenvalue for A if and only if it solves det ( A λI n ) = 0. (Here I n represents the n × n identity matrix.) (6 pts) b. For
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 4 1 − 2 A = 16 − 2 − 8 ⎞ ⎝ 4 ⎠ − 2 − 2 determine the eigenvalues. (2 pts) c. Use the results from part b to explain why A is not invertible. (2 pts) 5. Let X, Y be n × n matrices such that X 3 = Y 3 and X 2 Y = Y 2 X . What are nec-essary and sufficient conditions on X and Y such that X 2 + Y 2 is invertible? (4 pts) ± 1 MIT OpenCourseWare http://ocw.mit.edu 18.024 Multivariable Calculus with Theory Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern