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MATH1813 Problem Set 2 Ans

# MATH1813 Problem Set 2 Ans - University of Hong Kong Fall...

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University of Hong Kong Fall 2011 Math 1813 Mathematical Methods for Actuarial Science Instructor: Siye Wu Solution to Problem Set 2 I. (a) (i) True. This can be seen from the following λ is an eigenvalue of A ⇐⇒ det( A - λI n ) = 0 ⇐⇒ det( A T - λI n ) = 0 ⇐⇒ λ is an eigenvalue of A T . (ii) False. For example, let A = 0 1 0 0 , λ = 0 and v = 1 0 . Note that v is an eigenvector corresponding to the eigenvalue λ = 0 for A . The only eigenvalue of A T is 0 and v is not an eigenvector of A T . (b) True. If n is odd, then the characteristic polynomial det( A - λI n ) is a polynomial on λ of odd degree. Recalling the fact that a real polynomial of odd degree has at least one real root, we see the conclusion is true. (c) True. The reason is A is invertible ⇐⇒ the only solution to the system Ax = 0 is the zero solution ⇐⇒ zero is not an eigenvalue of A ⇐⇒ all the eigenvalues of A are non-zero . II. (a) Consider the characteristic equation det( A - λI 3 ) = - λ ( λ - 1)( λ + 1) = 0, we see the set of all eigenvalues of A is { 1 , 0 , - 1 } .

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MATH1813 Problem Set 2 Ans - University of Hong Kong Fall...

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