solhw3 - Solutions Homework 3 Numerical Linear Algebra 1...

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

View Full Document Right Arrow Icon
Solutions Homework 3 Numerical Linear Algebra 1 Fall 2010 Problem 3.1 Problem 3.1.6 Golub and Van Loan p. 93 Solution: This is a standard incremental algorithm based on a partitioning of the matrices involved. The algorithm is O ( n 2 ) over all when we start with k = n - 1, i.e., S c and T c are initialized as 1 × 1 matrices and getting the initial x c and w c is therefore O (1). If we can show that producing x + and w + requires O ( n - k ) computations then we have total work n - 1 k =1 O ( n - k ) = O ( n 2 ) as desired. The identities are easily verified through simple algebra by multiplying out the given partitioned linear system: σ u T 0 S c τ v T 0 T c γ x c - λγ λx c = β b c σ u T 0 S c τγ + v T x c T c x c - λγ λx c = β b c σ u T 0 S c τγ + v T x c w c - λγ λx c = β b c From this consider only the first equation to deduce γ = β - σv T x c - u T w c στ - λ as desired. An algorithm follows easily. Given w c and x c they can each be extended by one component to x + and w + η = v T x c μ = u T w c γ = ( β - σv T x c - u T w c ) / ( στ - λ ) ω = τγ + η and we have w + = ω w c and x + = γ x c Note that the computation of η and μ require 2( n - k ) + O (1) each and everything else in the basic step is O (1). We therefore have that x + and w + can be produced in O ( n - k ) as desired. 1
Image of page 1

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

View Full Document Right Arrow Icon
Problem 3.2 Recall that any unit lower triangular matrix L n × n can be written in factored form as L = M 1 M 2 · · · M n - 1 (1) where M i = I + l i e T i is an elementary unit lower triangular matrix (column form). Given the ordering of the elementary matrices, this factorization did not require any computation. Consider a simpler elementary unit lower triangular matrix (element form) that differs from the identity in one off-diagonal element in the strict lower triangular part, i.e., E ij = I + λ ij e i e T j where i = j . 3.2.a . Show that computing the product of two element form elementary matrices is simply superposition of the elements into the product given by E ij E rs = I + λ ij e i e T j + λ rs e r e T s whenever j = r . 3.2.b . Show that if j = r and i = s then computing E ij E rs with requires no compu- tation and E ij E rs = E rs E ij i.e., the matrices commute. 3.2.c . Write a column form elementary matrix M i in terms of element form elementary matrices. Does the order of the E ji matter in this product? 3.2.d . Show how it follows that the factorization of (1) is easily expressed in terms of element form elementary matrices. 3.2.e . Show that the expression from part (3.2.d) can be rearranged to form L = R 2 . . . R n where R i = I + e i r T i is an elementary unit lower triangular matrix in row form. Solution: Consider the following products: E ij E rs = I + λ ij e i e T j + λ rs e r e T s + λ ij λ rs e i ( e T j e r ) e T s (2) E rs E ij = I + λ ij e i e T j + λ rs e r e T s + λ ij λ rs e r ( e T s e i ) e T j . (3) If j = r then (2) says that E ij E rs requires no computation. Similarly, if s = i then (3) says that E rs E ij requires no computation. If j = r and s = i then clearly E ij E rs = E rs E ij = I + λ ij e i e T j + λ rs e r e T s .
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