hw5_2011_04_28_01_solutions

hw5_2011_04_28_01_solutions - EE263 S. Lall 2011.04.28.01...

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Unformatted text preview: EE263 S. Lall 2011.04.28.01 Homework 5 Solutions Due Thursday 5/5, at 5 PM on the second floor of Packard. No extensions permitted. We will post the solutions Thursday evening. 1. Computing the SVD by hand. 0235 For this problem, you can use Matlab to calculate matrix multiplications or find eigen- values or eigenvectors, but not to calculate the SVD directly. Consider the matrix A = bracketleftbigg 2 11 10 5 bracketrightbigg (a) Determine an SVD of A in the usual form A = U V T . The SVD is not unique, so find the one that has the minimal number of minus signs in U and V . (b) List the singular values, left singular vectors, and right singular vectors of A . Draw a careful, labeled picture of the unit ball in R 2 and its image under A , together with the singular vectors, with the coordinates of their vertices marked. (c) What is the matrix norm bardbl A bardbl and the Frobenius norm bardbl A bardbl F ? (d) Find A- 1 , not directly, but via the SVD. (e) Verify that det A = 1 2 and | det A | = 1 2 (f) What is the area of the ellipsoid onto which A maps the unit ball of R 2 ? Solution. (a) The singular values of A are the non-zero eigenvalues of AA T , which is A = bracketleftbigg 2 11 10 5 bracketrightbigg Solving for the eigenvalues as usual gives 1 = 200 2 = 50 The corresponding eigenvectors u 1 = bracketleftBigg 1 2 1 2 bracketrightBigg u 2 = bracketleftBigg 1 2- 1 2 bracketrightBigg are the left singular vectors of A . Similarly the right singular vectors are the eigenvectors of A T A , which are v 1 = bracketleftbigg- 3 5 4 5 bracketrightbigg v 2 = bracketleftbigg 4 5 3 5 bracketrightbigg So the SVD of A is A = bracketleftBigg 1 2 1 2 1 2- 1 2 bracketrightBigg bracketleftbigg 10 2 5 2 bracketrightbiggbracketleftbigg- 3 5 4 5 4 5 3 5 bracketrightbigg . (b) The plots are shown below. 1 EE263 S. Lall 2011.04.28.01-20-10 10 20-15-10-5 5 10 15 1 u 1 =(10,10) 2 u 2 =(-5,5) Left singular vectors-1-0.5 0.5 1-1-0.8-0.6-0.4-0.2 0.2 0.4 0.6 0.8 1 v 1 =(-0.6,0.8) v 2 =(-0.8,-0.6) Right singular vectors (c) We have bardbl A bardbl = 10 2 bardbl A bardbl F = 5 10 (d) The inverse of A is A- 1 = V - 1 U T which is A- 1 = bracketleftbigg...
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This note was uploaded on 07/28/2011 for the course EE 263 taught by Professor Boyd,s during the Summer '08 term at Stanford.

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hw5_2011_04_28_01_solutions - EE263 S. Lall 2011.04.28.01...

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