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Unformatted text preview: L. Vandenberghe 11/10/10 EE103 Homework 6 solutions 1. Exercise 9.10. (a) Ax = 0 means x 1 + 2 x 2 = 0 , 3 x 1 6 x 2 = 0 , 2 x 1 x 2 = 0 . The first and second equations are satisfied if x 1 = 2 x 2 . The third equation is satisfied if 2 x 1 = x 2 . This is only possible if x 1 and x 2 are both zero. Therefore A has a zero nullspace. (b) Only m nmatrices with m n can have a zero nullspace. For this matrix, we can note that x = (3 , 1 , 0) is in the nullspace. (c) We have Ax = bracketleftBigg D B bracketrightBigg x = bracketleftBigg Dx Bx bracketrightBigg . Ax = 0 means Dx = 0 and Bx = 0. D is diagonal with nonzero diagonal elements, so Dx = 0 is only possible if x is zero. We conclude that A has a zero nullspace, regardless of the properties of B . (d) We have Ax = ( I U ) x = x Ux, so Ax = 0 means x = Ux . This is not possible for nonzero x because bardbl Ux bardbl bardbl U bardblbardbl x bardbl &lt; bardbl x bardbl if bardbl U bardbl &lt; 1 and x negationslash = 0. Therefore A has a zero nullspace. 2. Exercise 9.13. (a) The problem is equivalent to a leastsquares problem minimize vextenddouble vextenddouble vextenddouble vextenddouble vextenddouble bracketleftBigg A I bracketrightBigg x bracketleftBigg b bracketrightBiggvextenddouble vextenddouble vextenddouble vextenddouble vextenddouble 2 . The solution is x = bracketleftBigg A I bracketrightBigg T bracketleftBigg A I bracketrightBigg  1 bracketleftBigg A I bracketrightBigg T bracketleftBigg b bracketrightBigg = ( A T A + I ) 1 A T b. 1 (b) The identity A T ( AA T + I ) 1 = ( I + A T A ) 1 A T can be verified by multiplying both sides of the equation on the left with I + A T A and on the right with AA T + I . On the lefthand side we get ( I + A T A ) A T = A T +...
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This note was uploaded on 02/03/2011 for the course EE 103 taught by Professor Vandenberghe,lieven during the Spring '08 term at UCLA.
 Spring '08
 VANDENBERGHE,LIEVEN

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