pset8-soln

pset8-soln - 18.06 Problem Set 8 Due Wednesday 23 April...

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Unformatted text preview: 18.06 Problem Set 8 Due Wednesday, 23 April 2008 at 4 pm in 2-106. Problem 1: Do problem 3 in section 6.5 (pg. 339) in the book. Solution (10 points) The matrix A encodes the quadratic f = x 2 +4 xy +9 y 2 . Completing the square is the same as finding the row reduced form of A : we find U = 1 2 0 5 = 1 0 0 5 1 2 0 1 so f = ( x + 2 y ) 2 + 5 y 2 . Similarly, B gives the quadratic f = x 2 +6 xy +9 y 2 . This time the reduced form is U = 1 3 0 0 = 1 0 0 0 1 3 0 1 so f = ( x + 3 y ) 2 . Problem 2: Do problem 6 in section 6.5 (pg. 339). Solution (10 points) If A has full column rank, we know that A T A is square symmetric and invertible. We must show that it is also positive definite. The easiest criterion is to show that for every non-zero vector x the number x T ( A T A ) x > 0. To do this we just note that x T A T Ax = Ax · Ax = k Ax k 2 > 0. Problem 3: For what numbers c and d are the matrices A and B positive definite? Test the 3 determinants: A = c 2 3 2 c 4 3 4 1 B = 1 2 1 2 d 3 1 3 1 Solution (15 points) 1 The three top-left determinants of A are c , c 2- 4, and c ( c- 16)- 2(- 10) + 3(8- 3 c ) = c 2- 25 c + 44. If A is positive definite, all three of these numbers must be positive. First of all c > 0. The second determinant yields c 2- 4 = ( c- 2)( c + 2) > so either c > 2 or c <- 2. Because c > 0 we can ignore the second case. Finally, the third determinant c 2- 25 c +44 has roots 25 / 2 ± 1 2 √ 449 and we either need c to be smaller than the smaller root or larger than the larger root. The smaller root is less than 2, so we can ignore that piece of it. In the end we find c > 25 + √ 449 2 The top three determinants of B are 1, d- 4, and ( d- 9)- 2(- 1)+1(6- d ) =- 1. This last determinant is negative no matter what d is, so this matrix will never be positive definite. Problem 4: Do problem 15 in section 6.5 (pg. 340). Solution (10 points) We must show that if A and B are positive definite, so is A + B . We use the x T ( A + B ) x criterion: we have x T ( A + B ) x = x T Ax + x T Bx . If x is nonzero than both terms are positive (since A and B are positive definite), so the whole thing is positive. Thus A + B is also positive definite....
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This note was uploaded on 08/10/2008 for the course MATH 250 taught by Professor Chanillo during the Spring '08 term at Rutgers.

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pset8-soln - 18.06 Problem Set 8 Due Wednesday 23 April...

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