q2_sp12_sol - 18.06 Professor Strang Quiz 2 April 11th 2012...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
18.06 Professor Strang Quiz 2 April 11th, 2012 Grading 1 2 3 Your PRINTED name is: Please circle your recitation: r01 T 11 4-159 Ailsa Keating ailsa r02 T 11 36-153 Rune Haugseng haugseng r03 T 12 4-159 Jennifer Park jmypark r04 T 12 36-153 Rune Haugseng haugseng r05 T 1 4-153 Dimiter Ostrev ostrev r06 T 1 4-159 Uhi Rinn Suh ursuh r07 T 1 66-144 Ailsa Keating ailsa r08 T 2 66-144 Niels Martin Moller moller r09 T 2 4-153 Dimiter Ostrev ostrev r10 ESG Gabrielle Stoy gstoy
Image of page 1

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

View Full Document Right Arrow Icon
1 (40 pts.) (a) Find the projection p of the vector b onto the plane of a 1 and a 2 , when b = 1 0 0 1 , a 1 = 1 7 1 7 , a 2 = - 1 7 1 - 7 . Solution. Observe that a T 1 a 2 = 0 . Thus p = a T 1 b a T 1 a 1 a 1 + a T 2 b a T 2 a 2 a 2 = 8 100 a 1 - 8 100 a 2 = 4 / 25 0 0 28 / 25 . (b) What projection matrix P will produce the projection p = Pb for every vector b in R 4 ? Solution. Let A be the 4 × 2 matrix with columns a 1 , a 2 . P is given by P = A ( A T A ) - 1 A T . Notice that A T A = 100 0 0 100 . ( a 1 and a 2 are orthogonal and of same length.) Thus P = 1 100 AA T = 1 100 2 0 0 14 0 98 14 0 0 14 2 0 14 0 0 98 . Page 2 of 9
Image of page 2
(c) What is the determinant of I - P ? Explain your answer. Solution. I - P is the matrix of the projection to the orthgonal complement of C ( A ) , i.e. N ( A T ) . In particular, I - P has rank the dimension of N ( A T ) , which is 3. Thus I - P is singular, and det( I - P ) = 0 .
Image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
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