20105ee205A_1_2010ee205A_1_HW4_sol

20105ee205A_1_2010ee205A_1_HW4_sol - Chapter 4 An...

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

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20105ee205A_1_2010ee205A_1_HW4_sol - Chapter 4 An...

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