20095ee103_1_hw6_sols

# 20095ee103_1_hw6_sols - L Vandenberghe EE103 Homework 6...

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Unformatted text preview: L. Vandenberghe 11/12/09 EE103 Homework 6 solutions 1. Exercise 9.9. The first plot is for μ = 1, the second plot is for μ = 100, the last plot is for μ = 10000. For μ = 1, there is not enough smoothing, for μ = 10000 we have too much smoothing, and μ = 100 seems about right. 200 400 600 800 1000-0.5 0.5 200 400 600 800 1000-0.5 0.5 200 400 600 800 1000-0.5 0.5 i i ˆ x ˆ x ˆ x 2. Exercise 9.11. (a) We use the property that a square matrix B is nonsingular if and only if Bx = 0 implies x = 0. Suppose x ∈ R m and y ∈ R n satisfy bracketleftBigg I A A T bracketrightBigg bracketleftBigg x y bracketrightBigg = 0 . In other words, x + Ay = 0 and A T x = 0. Then A T x = − A T Ay = 0 . A T A is positive definite (hence, nonsingular) if A has a zero nullspace. Therefore y = 0, and from x = − Ay we can also conclude that x = 0. 1 (b) The equations bracketleftBigg I A A T bracketrightBigg bracketleftBigg x y bracketrightBigg = bracketleftBigg b bracketrightBigg are equivalent to x + Ay = b, A T x = 0 . We have to check that those two equations are satisfied for x = b − Ax ls , y = x ls . Clearly this choice of x and y satisfies x + Ay = b . Moreover, we have A T x = A T ( b − A ( A T A )- 1 A T b ) = A T b − ( A T A )( A T A )- 1 A T b = A T b − A T b = 0 . 3. Exercise 10.4. (a) The second problem is a least-squares problem minimize vextenddouble vextenddouble vextenddouble vextenddouble vextenddouble bracketleftBigg A c T bracketrightBigg y − bracketleftBigg b d bracketrightBiggvextenddouble vextenddouble vextenddouble vextenddouble vextenddouble . We need to verify that ˆ y satisfies the normal equations ( A T A + cc T )ˆ y = A T b + dc. We know that A T A ˆ x = A T b , so ( A T A + cc T )ˆ x = A T b + ( c T ˆ x ) c. Also, ( A T A + cc T )( A T A )- 1 c = c + ( c T ( A T A )- 1 c ) c. Therefore ( A T A + cc T )ˆ y = ( A T A + cc T ) parenleftBigg ˆ x + d − c T ˆ x 1 + c T ( A T A )- 1 c ( A T A )- 1 c parenrightBigg = A T b + ( c T ˆ x ) c + d − c T ˆ x 1 + c T ( A T A )- 1 c (1 + c T ( A T A )- 1 c ) c = A T b + ( c T ˆ x ) c + ( d − c T ˆ x ) c = A T b + dc. 2 (b) If we plug in the QR factorization of A , the formulas for ˆ x and ˆ y reduce to ˆ x = R- 1 Q T b ˆ y = ˆ x + d − c T ˆ x 1 + c T R- 1 R- T c R- 1 R- T c. • QR factorization of A (2 mn 2 flops). • Compute u = Q T b (2 mn flops). • Compute ˆ x = R- 1 u using backward substitution ( n 2 flops). • Compute w = R- 1 R- T c using forward and backward substitution (2 n 2 flops). • Compute ˆ y = ˆ x + d- c T ˆ x 1+ c T w w (6 n flops). The total is 2 mn 2 + 2 mn + 3 n 2 ....
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20095ee103_1_hw6_sols - L Vandenberghe EE103 Homework 6...

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