20105ee205A_1_2010ee205A_1_HW4_sol

# 20105ee205A_1_2010ee205A_1_HW4_sol - Chapter 4 An...

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

Chapter 4 An Introduction to the Moore-Penrose Pseudoinverse 1. Use Theorem 4.4 to compute the pseudoinverse of ± 11 22 ² . Answer 4.1 A + = lim δ 0 ( A T A + δ 2 I ) - 1 A T = lim δ 0 ³± 12 ²± ² + ± δ 2 0 0 δ 2 ²´ - 1 ± ² = lim δ 0 ± 5+ δ 2 5 5 5 + δ 2 ² - 1 ± ² = lim δ 0 1 (5 + δ 2 ) 2 - 5 2 ± δ 2 - 5 - 5 5 + δ 2 ² = lim δ 0 1 10 δ 2 + δ 4 ± δ 2 2 δ 2 δ 2 2 δ 2 ² = lim δ 0 1 10 + δ 2 ± ² 13

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

View Full Document
14 CHAPTER 4. AN INTRODUCTION TO THE MOORE-PENROSE PSEUDOINVERSE = 1 10 ± 12 ² . 2. If x,y IR n , show that ( xy T ) + =( x T x ) + ( y T y ) + yx T . Answer 4.2 Let A = xy T and G x T x ) + ( y T y ) + T . We want to show that G = A + . This is true if and only if the four Penrose conditions are satisFed. If x =0 or y , then A = 0 = G = A + , so suppose ± . Then AGA =[ xy T ][( x T x ) + ( y T y ) + T ][ xy T ] x T x ) - 1 ( y T y ) - 1 x ( y T y )( x T x ) y T = xy T = A. GAG = [( x T x ) + ( y T y ) + T ][ xy T ][( x T x ) + ( y T y ) + T ] x T x ) + ( y T y ) + y ( x T x )( y T y ) x T ( x T x ) - 1 ( y T y ) - 1 x T x ) + ( y T y ) + T = G. ( AG ) T = ([ xy T ][( x T x ) + ( y T y ) + T ]) T = [( x T x ) + ( y T y ) + T ] T [ xy T ] T = xy T ( x T x ) + ( y T y ) + T = AG. ( GA ) T = ([( x T x ) + ( y T y ) + T ][ xy T ]) T xy T ] T [( x T x ) + ( y T y ) + T ] T x T x ) + ( y T y ) + T xy T = GA. 3. For A m × n , prove that R ( A )= R ( AA T ) using only de±nitions and elementary properties of the Moore-Penrose pseudoinverse. Answer 4.3 Let x ∈ R ( A ) . Then there exists v n such that x = Av = AA + Av = A ( A + A ) T v = AA T ( A + ) T v ( AA T ⇒ R ( A ) R ( AA T ) . Conversely, x ( AA T there exists v m such that x = AA T v ( A ( AA T ) ⊆ R ( A ) . Thus R ( A R ( AA T ) .
15 4. For A IR m × n , prove that R ( A + )= R ( A T ). Answer 4.4 Let x ∈ R ( A + ) . Then there exists v n such that x = A + v = A T ( AA T ) + v ( A T ) . Therefore R ( A + ) ⊆ R ( A T ) . Conversely, x ( A T there exists v m such that x = A T v = ( AA + A ) T v =( A + A ) T A T v = A + AA T v ( A + ) . Therefore R ( A T ) R ( A + ) . Thus R ( A + R ( A T ) . 5. For A p × n and B m × n show that N ( A ) ⊆ N ( B ) if and only if BA + A = B . Answer 4.5 N ( A ) ( B ) ⇐⇒ N ( B ) ( A ) R ( B T ) ( A T ) B T A T )( A T ) + B T B T = A T ( A + ) T B T B = + A.

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.

## This note was uploaded on 04/18/2011 for the course EE 205A taught by Professor Laub during the Fall '10 term at UCLA.

### Page1 / 8

20105ee205A_1_2010ee205A_1_HW4_sol - Chapter 4 An...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online