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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 1 1 2 2 . Answer 4.1 A + = lim δ 0 ( A T A + δ 2 I ) - 1 A T = lim δ 0 1 2 1 2 1 1 2 2 + δ 2 0 0 δ 2 - 1 1 2 1 2 = lim δ 0 5 + δ 2 5 5 5 + δ 2 - 1 1 2 1 2 = lim δ 0 1 (5 + δ 2 ) 2 - 5 2 5 + δ 2 - 5 - 5 5 + δ 2 1 2 1 2 = lim δ 0 1 10 δ 2 + δ 4 δ 2 2 δ 2 δ 2 2 δ 2 = lim δ 0 1 10 + δ 2 1 2 1 2 13

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

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20105ee205A_1_2010ee205A_1_HW4_sol - Chapter 4 An...

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