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MIT18_06S10_pset9_s10_soln

# MIT18_06S10_pset9_s10_soln - 18.06 Pset 9 Solutions...

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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 ) . 0 2 8 25 Solution (4 points) From C , we obtain � � A = C T C = 3 0 3 1 = 9 3 . 1 2 0 2 3 5 4 0 1 0 Conversely, the given A quickly diagonalizes to via L = : thus 0 9 2 1 2 4 C = chol ( A ) = DL T = . 0 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 0 0 1 1 1 A = 0 1 2 and A = 1 2 2 0 2 8 1 2 7 . Solution (4 points) For the first matrix A , we have ⎤ ⎡ ⎤ ⎡ 1 0 0 9 0 0 1 0 0 3 0 0 A = 0 1 0 ⎦ ⎣ 0 1 0 ⎦ ⎣ 0 1 2 C = 0 1 0 0 2 1 0 0 4 0 0 1 0 2 2 while for the second matrix we have ⎤ ⎡ ⎤ ⎡ 1 0 0 1 0 0 1 1 1 1 1 1 A = 1 1 0 ⎦ ⎣ 0 1 0 ⎦ ⎣ 0 1 1 C = 0 1 1 5 1 1 1 0 0 5 0 0 1 0 0 1

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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 0 a 0 1 b/a x x y = x y c d y b/a 1 0 ( ac b 2 ) /a 0 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). Solution (4 points) For F 1 ( x, y ), we first solve for the stationary point ∂F 1 ∂F 1 = x 3 + 2 x + 2 xy = 0 , = x 2 + 2 y = 0 ∂x ∂y From (2), we have y = x 2 / 2. Plug this into (1), we have 2 x = 0 and hence the only critical point is x =
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