Scientific Computing

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

View Full Document Right Arrow Icon
CS205 Homework #7 Solutions Problem 1 We have seen the application of the conjugate gradient algorithm on the solution of sym- metric, positive definite systems. Now assume that in the system Ax = b , the n × n matrix A is symmetric positive semi-definite with a nullspace of dimension p < n . This problem illustrates that one can use a modified version of conjugate gradients to solve this system as well. 1. Prove that we can write A as A = M ˜ AM T where M is an n × ( n - p ) matrix with orthonormal columns that form a basis for the column space of A , while ˜ A is an ( n - p ) × ( n - p ) symmetric positive definite matrix (no nullspace) [Hint: Use the diagonal form of A = QΛQ T ] 2. Let the n × n matrix P be defined as P = MM T . Explain (no formal proof required) why this is a projection matrix and onto what space it projects. How can we compute P without knowledge of the eigenvalues-eigenvectors of A ? 3. Show that, in order to have a solution to Ax = b , we must be able to write b = M ˜ b for an appropriate vector ˜ b R n - p 4. Let ˜x be the solution to the system ˜ A˜x = ˜ b and explain why ˜x is unique. Show that any solution to the original system Ax = b can be written as x = M˜x + x 0 where x 0 is in the nullspace of A . 5. Consider the conjugate gradients algorithm for solving ˜ A˜x = ˜ b ˜x 0 = initial guess ˜ s 0 = ˜ r 0 = ˜ b - ˜ A˜x 0 for k = 0 , 1 , . . . , 2 ˜ α k = ˜ r T k ˜ r k ˜ s T k ˜ s k ˜x k +1 = ˜x k + ˜ α k ˜ s k ˜ r k +1 = ˜ r k - ˜ α k ˜ s k ˜ s k +1 = ˜ r k +1 + ˜ r T k +1 ˜ r k +1 ˜ r T k ˜ r k ˜ s k end 1
Image of page 1

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

View Full Document Right Arrow Icon
Show that we can compute a solution to the original system Ax = b by using the following modification of the algorithm x 0 = initial guess s 0 = r 0 = P ( b - Ax 0 ) for k = 0 , 1 , . . . , 2 α k = r T k r k s T k As k x k +1 = x k + α k s k r k +1 = r k - α k PAs k s k +1 = r k +1 + r T k +1 r k +1 r T k r k s k end [Hint: Show that x k = M˜x k , r k = r k , s k = s k , ˜ α k = α k ] Solution 1. Since A is symmetric and positive definite it can be written as A = QΛQ T = q 1 q 2 · · · q n λ 1 λ 2 . . . λ n q T 1 q T 2 .
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