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Unformatted text preview: L. Vandenberghe 12/09/10 EE103 Final Exam Solutions Problem 1 (20 points). A matrix of the form P = I − 2 u T u uu T , where u is a nonzero n-vector, is called a Householder matrix of order n . 1. (7 points) Show that P is orthogonal. 2. (7 points) Is P positive definite? 3. (6 points) Suppose you are given the vector u and an n × m-matrix A . How would you compute the matrix-matrix product PA as efficiently as possible? What is the cost of your method (number of flops as a function of m and n )? Solution. 1. P is symmetric, so we need to verify that the product of P with itself is the identity: P 2 = ( I − 2 u T u uu T )( I − 2 u T u uu T ) = I − 2 u T u uu T − 2 u T uu T + 4 ( u T u ) 2 uu T uu T = I − 4 u T u uu T + 4( u T u ) ( u T u ) 2 uu T = I − 4 u T u uu T + 4 u T u uu T = I. 2. P is not positive definite. To prove this, it is sufficient to find a nonzero x with x T Px ≤ 0. For example, if we take x = u , then u T Pu = u T u − 2 u T u ( u T u ) 2 = −bardbl u bardbl 2 < . 3. We write the product as PA = A − 2 u T u u ( A T u ) T . Computing γ = u T u and w = (2 /γ ) u takes 3 n flops. The product v = A T u takes 2 mn flops. The product C = wv T takes mn flops and subtracting it from A another mn flops. The total is 4 mn . This is more efficient than first computing P and multiplying with A , which costs 2 n 2 m flops. 1 Problem 2 (20 points). Suppose Q is square and orthogonal. For each of the following matri- ces A , give bardbl A bardbl and, if A is invertible, also bardbl A − 1 bardbl . Explain your answers. 1. (10 points) A = bracketleftBigg Q − Q Q Q bracketrightBigg . 2. (10 points) A = bracketleftBigg Q Q Q Q bracketrightBigg . Solution. If Q is square and orthogonal, then Q T Q = QQ T = I . Also bardbl Q bardbl = bardbl Q T bardbl = 1 and bardbl Qy bardbl = bardbl Q T y bardbl = bardbl y bardbl for all y . 1. We note that (1 / √ 2) A is orthogonal: 1 2 bracketleftBigg Q T Q T − Q T Q T bracketrightBiggbracketleftBigg Q − Q Q Q bracketrightBigg = 1 2 bracketleftBigg 2 I 2 I bracketrightBigg . Orthogonal matrices have norm 1; therefore bardbl A bardbl = √ 2. The same identity shows that A − 1 = (1 / 2) A T and that √ 2 A − 1 is orthogonal. Therefore bardbl A − 1 bardbl = 1 / √ 2. 2. We partition the vector x in the definition bardbl A bardbl = max x negationslash =0 bardbl Ax bardbl / bardbl x bardbl x = bracketleftBigg u v bracketrightBigg , where u and v are n-vectors. Note that bardbl x bardbl = ( bardbl u bardbl 2 + bardbl v bardbl 2 ) 1 / 2 . We have bardbl Ax bardbl = vextenddouble vextenddouble vextenddouble vextenddouble vextenddouble bracketleftBigg Q ( u + v ) Q ( u + v ) bracketrightBiggvextenddouble vextenddouble vextenddouble vextenddouble vextenddouble = parenleftBig 2 bardbl Q ( u + v ) bardbl 2 parenrightBig 1 / 2 = √ 2 bardbl u + v bardbl ....
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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