고려대 경제수학 2013 중간 기말 답지.pdf

Econ 205 Final, Fall Semester 2013, 09:00 10:15 [18 points] 1. Eigenvalues and Eigenvectors of an identity matrix. not an eigenvalue. [4] (3) The condition we need is (A eigenvalues: det (A rI) = (1 ! v1 = [5] (4) Av = A v2 rI) is not invertible. Therefore, use det (A rI) = 0 to nd r) (2 r) !4 = (r + 3) ! 2). We nd two eigenvalues: 3 and 2. (r v1 v1 + 2v2 . From v1 + 2v2 = 2v1 , we have v1 = 2v2 . = 2 v2 2v1 2v2 2 1 p ;p 5 5 T [3] (1) Av = rv , Av = rIv , (A r0 I) 1 rI) v = 0, where the second equation holds due to the property r0 I) v = 0 and we get v = 0. Since v = 0, r0 is [6] (2) Premultiply (A to both sides of (A Therefore, (2; 1)T is an eigenvector that corresponds to the eigenvalue 2. The eigenvector that satises the conditions in the problem: . [12 points] 2. Unconstrained Optimization [3] (1) As f is a C 2 function, f 00 exists and is continuous. So, we have limh!0 f 00 (x + h) = f 00 (x). Therefore, f 0 (x0) h + R2 (h) h2 = 1 2 [f 00 (x + h) [6] (2) Note that f 0…
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