Linear Algebra with Applications (3rd Edition)

Info icon This preview shows pages 1–3. Sign up to view the full content.

Math 417 homework 2 solutions (Given only for problems that are not straightforward computation; contact the instructor if you still have questions about the others.) Section 2.1 problem 1 A transformation T : R 3 R 3 is linear only if it satisfies T (0) = 0. The transformation is not linear. For x 1 = x 2 = x 3 = 0 we get y 2 = 2, not y 0 = 0. However, the transformation can be written T ( x ) = 0 2 0 0 1 0 0 2 0 x 1 x 2 x 3 + 0 2 0 . Outside of algebra, adding a constant vector still counts as “linear”. Section 2.1 problem 1 A linear transformation T : R 3 R 3 must satisfy T ( α · x ) = α · x for all α R . But here 2 T ( 1 0 1 ) = 2 - 1 1 1 = - 2 2 2 which is not the same as T (2 1 0 1 ) = T ( 2 0 2 ) = - 2 4 2 . Note: alternatively one could check that another property of linear transfor- mations, T ( x + y ) = T ( x ) + T ( y ) , is not satisfied either. 1
Image of page 1

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

Section 2.1 problem 44 The transformation is linear, with matrix 0 - a 3 a 2 a 3 0 - a 1 - a 2 a 1 0 . Section 2.2 problem 2 Counterclockwise rotation around the origin by an angle α is multiplication by the matrix cos α - sin α sin α cos α For α = 60 we get 1 2 - 3 2 3 2 1 2 Section 2.2 problem 10 4 3 is a vector on the line. Its length is 4 2 + 3 2 = 25 = 5, so a unit vector on the line is u = u 1 u 2 = 4 / 5 3 / 5 . The projection matrix is u 2 1 u 1 u 2 u 1 u 2 u 2 2 = 16 / 25 12 / 25 12 / 25 9 / 25 = 1 25 16 12 12 9 . The reflection matrix is 2 u 2 1 - 1 2 u 1 u 2 2 u 1 u 2 2 u 2 2 - 1 = 7 / 25 24 / 25 24 / 25 - 7 / 25 = 1 25 7 24 24 - 7 .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the 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