A8_soln

# A8_soln - Math 235 Assignment 8 Solutions 1. Find a...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 235 Assignment 8 Solutions 1. Find a singular value decomposition of each of the following matrices. a)- 2- 1 Solution: Let A =- 2- 1 then A T A = 4 0 0 1 . The eigenvalues of A T A are λ 1 = 4 and λ 2 = 1 and A T A is already diagonal so we take V = 1 0 0 1 . The singular values of A are σ 1 = √ 4 = 2 and σ 2 = √ 1 = 1. Thus Σ = 2 0 0 1 . Next compute ~u 1 = 1 σ 1 A~v 1 = 1 2- 2- 1 1 =- 1 ~u 2 = 1 σ 2 A~v 2 = 1 1- 2- 1 1 =- 1 Hence we take U =- 1- 1 and get that A = U Σ V T as required. b) 4- 2 2- 1 Solution: Let A = 4- 2 2- 1 then A T A = 20- 10- 10 5 . The eigenvalues of A T A are λ 1 = 25 and λ 2 = 0 with corresponding normalized eigenvectors ~v 1 = 2 / √ 5- 1 / √ 5 and ~v 2 = 1 / √ 5 2 / √ 5 . Thus we take V = 2 / √ 5 1 / √ 5- 1 / √ 5 2 / √ 5 . The singular values of A are σ 1 = √ 25 = 5 and σ 2 = √ 0 = 0. Thus Σ = 5 0 0 0 0 0 . Next compute ~u 1 = 1 σ 1 A~v 1 = 1 5 4- 2 2- 1 2 / √ 5- 1 / √ 5 = 2 / √ 5 1 / √ 5 We only have one vector, ~u 1 , in R 3 , so we must extend this to form an orthonormal basis for R 3 . Observe that we just need to pick two vectors that are orthogonal to ~u 1 and to each other. That is vectors x 1 x 2 x 3 such that 2 / √ 5 1 / √ 5 · x 1 x 2 x 3 = 0 . This gives the system 2 x 1 + x 2 = 0. Hence, we pick (1 ,- 2 , 0) and (0 , , 1) which are clearly orthogonal to each 2 other. Normalizing them gives ~u 2 = 1 / √ 5- 2 / √ 5 , ~u 3 = 1 and so we take U = ~u 1 ~u 2 ~u 3 . Then A = U Σ V T as required. c) 1 1 1- 1 1 Solution: Let A = 1 1 1- 1 1 then A T A = 2 0 0 3 and the eigenvalues of A T A are seen to be (in decreasing order) λ 1 = 3 and λ 2 = 2. Corresponding normalized eigenvectors are ~v 1 = 1 for λ 1 and ~v 2 = 1 for λ 2 . Thus one choice for V is V = 0 1 1 0 . The singular values of A are σ 1 = √ 3 and σ 2 = √ 2. Thus Σ = √ 3 √ 2 Next compute ~u 1 = 1 σ 1 A~v 1 = 1 √ 3 1 1 1- 1 1 1 = 1 / √ 3 1 / √ 3 1 / √ 3 ~u 2 = 1 σ...
View Full Document

## This note was uploaded on 04/18/2010 for the course MATH 235 taught by Professor Celmin during the Spring '08 term at Waterloo.

### Page1 / 6

A8_soln - Math 235 Assignment 8 Solutions 1. Find a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online