20105ee205A_1_2010ee205A_1_HW5_sol

# 20105ee205A_1_2010ee205A_1_HW5_sol - Chapter 6 Solution of...

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Chapter 6 Solution of Linear Equations 1. As in Example 6.8, characterize all left inverses of a matrix A IR m × n . Answer 6.1 A has a left inverse ⇐⇒ A T has a right inverse R ( I n ) ⊆ R ( A T ) A T ( A T ) + I n = I n rank( A T )= r = n (since r n ) A T is onto ( ( A T ) + is then a right inverse, so A + is a left inverse). Thus, all left inverses of A are of the form L = ± ( A T ) + I n +( I m - ( A T ) + A T ) Z T ² T = ± ( A T ) + I - ( A T ) + A T ) Z T ² T = A + + Z ( I - AA + ) where Z n × m is arbitrary. 2. Let A m × n , B m × k and suppose A has an SVD as in Theorem 5.1. Assuming R ( B ) ( A ), characterize all solutions of the matrix linear equation AX = B in terms of the SVD of A . X = A + B I - A + A ) Y = V 1 S + U T 1 B I - V 1 V T 1 ) Y = V 1 S - 1 U T 1 B + V 2 V T 2 Y where Y n × k is arbitrary. 21

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22 CHAPTER 6. SOLUTION OF LINEAR EQUATIONS 3. Let x,y IR n and suppose further that x T y ± = 1. Show that ( I - xy T ) - 1 = I - 1 x T y - 1 xy T . Answer 6.2 By the Sherman-Morrison-Woodbury formula, ( I - xy T ) - 1 =( I + x ( - 1) y T ) - 1 = I - Ix ( - 1+ y T ) - 1 y T I = I - x ( y T x - 1) - 1 y T = I - 1 x T y - 1 xy T . 4. Let n and suppose further that x T y ± = 1. Show that ± y T 1 ² - 1 = ± I + cxy T - cx - cy T c ² where c =1 / (1 - x T y ). Answer 6.3 By formula 7 in Section 6.4, ± y T 1 ² - 1 = ± I + Ixcy T I - Ixc - cy T Ic ² = ± I + cxy T - cx - cy T c ² where c = 1 1 - y T = 1 1 - x T y . 5. Let A n × n n and let A - 1 have columns c 1 ,...,c n and individual elements γ ij . Assuming that γ ji ± = 0, show that the matrix B = A - 1 γ e i e T j (i.e., A with 1 γ subtracted from its ij -th element) is singular. Hint: Show that c i ∈ N ( B ). Answer 6.4 If we show that N ( B ) ± =0 , then this will mean that B is singular since it will not have full rank. Now Bc i = Ac i - 1 γ e i e T j c i = e i - 1 γ e i γ = e i - e i . Hence, c i ( B ) . But c i ± since A is nonsingular. Therefore, B is singular.
23 6. As in Example 6.10, check directly that the condition for reconstructibil-

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## This note was uploaded on 04/19/2011 for the course EE 205A taught by Professor Laub during the Fall '10 term at UCLA.

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20105ee205A_1_2010ee205A_1_HW5_sol - Chapter 6 Solution of...

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