MIT18_06S10_pset9_s10_soln

MIT18_06S10_pset9_s1 - 18.06 Pset 9 Solutions Problem 6.5#25 With positive pivots in D the factorization A = LDL T becomes L √ D √ DL T(Square

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Unformatted text preview: 18.06 Pset 9 Solutions Problem 6.5, #25: With positive pivots in D , the factorization A = LDL T becomes L √ D √ DL T . (Square roots of the pivots give D = √ D √ D .) Then C = √ DL T yields the Cholesky factorization A = C T C which is “symmetrized LU ”. 3 1 4 8 From C = find A. From A = find C = chol ( A ) . 2 8 25 Solution (4 points) From C , we obtain A = C T C = 3 3 1 = 9 3 . 1 2 2 3 5 4 1 Conversely, the given A quickly diagonalizes to via L = : thus 9 2 1 2 4 C = chol ( A ) = √ DL T = . 3 Problem 6.5, #26: In the Cholesky factorization A = C T C , with C T = L √ D , the square roots of the pivots are on the diagonal of C . Find C (upper triangular) for ⎡ ⎤ ⎡ ⎤ 9 1 1 1 A = ⎣ 1 2 ⎦ and A = ⎣ 1 2 2 ⎦ 2 8 1 2 7 . Solution (4 points) For the first matrix A , we have ⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 9 1 3 A = ⎣ 1 ⎦⎣ 1 ⎦⎣ 1 2 ⎦ C = ⎣ 1 ⎦ ⇒ 2 1 4 1 2 2 while for the second matrix we have ⎡ ⎤⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 1 1 1 1 1 1 1 A = ⎣ 1 1 ⎦⎣ 1 ⎦⎣ 1 1 ⎦ C = ⎣ 1 1 ⎦ ⇒ √ 5 1 1 1 5 1 1 2 Problem 6.5, #27: The symmetric factorization A = LDL T means that x T A x = x T LDL T x : a b x 1 a 1 b/a x x y = x y c d y b/a 1 ( ac − b 2 ) /a 1 y The left side is ax 2 + 2 bxy + cy 2 . The right side is a ( x + a b y ) 2 + y 2 . The second pivot completes the square! Test with a = 2 ,b = 4 ,c = 10. Solution (4 points) Evaluating out the right side gives ax 2 + 2 bxy + cy 2 , so the b 2 entry in the space given is c − a , i.e. the second pivot. For the given values, we have 2 x 2 + 8 xy + 10 y 2 = 2( x + 2 y ) 2 + 2 y 2 as desired. Problem 6.5, #29: For F 1 ( x,y ) = x 4 / 4+ x 2 + x 2 y + y 2 and F 2 ( x,y ) = x 3 + xy − x , find the second-derivative matrices H 1 and H 2 : ∂ 2 F ∂ 2 F ∂x 2 ∂x∂y H = ∂ 2 F ∂ 2 F . ∂y∂x ∂y 2 H 1 is positive-definite so F 1 is concave up (= convex). Find the minimum point of F 1 and the the saddle point of F 2 (look only where the first derivatives are zero)....
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This note was uploaded on 08/22/2011 for the course MATH 1806 taught by Professor Strang during the Fall '10 term at MIT.

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MIT18_06S10_pset9_s1 - 18.06 Pset 9 Solutions Problem 6.5#25 With positive pivots in D the factorization A = LDL T becomes L √ D √ DL T(Square

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