Problem_Set_5_Solutions

# Problem_Set_5_Solutions - Problem Set 5 Solutions Economics...

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Problem Set 5 Solutions Economics 6170 TA: Christopher Handy October 17, 2008 Problem 1 (i) Note that p = tr( A ) = a 11 + a 22 and q = det A = a 11 a 22 - a 12 a 21 . Then we want to solve 0 = f ( λ ) = det( A - λI ) = ( a 11 - λ )( a 22 - λ ) - a 12 a 21 = λ 2 - ( a 11 + a 22 ) λ + ( a 11 a 22 - a 12 a 21 ) = λ 2 - + q By the quadratic formula, the roots are λ = p ± p p 2 - 4 q 2 We are given that p 2 > 4 q , so there are two distinct real roots. Without loss of generality, let’s call the larger root λ 1 . Then we have λ 1 = p + p p 2 - 4 q 2 , λ 2 = p - p p 2 - 4 q 2 , λ 1 > λ 2 Since p > 0, we have λ 1 > 0. The smaller root λ 2 is positive if and only if p - p p 2 - 4 q > 0 ⇐⇒ p > p p 2 - 4 q ⇐⇒ p 2 > p 2 - 4 q ⇐⇒ q > 0 where p > p p 2 - 4 q ⇐⇒ p 2 > p 2 - 4 q holds since p > 0. Because we are given q > 0, we have λ 2 > 0. 1

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(ii) If λ 1 = 1 or λ 2 = 1, then p - q = 1, which is a contradiction of the given information that p - q < 1. So, in the proof below we consider only strict inequalities. (A): λ 1 < 1 and λ 2 < 1. (not A): λ 1 > 1 or λ 2 > 1. (B): λ 1 > 1 and λ 2 > 1. (not B): λ 1 < 1 or λ 2 < 1. We want to show that (A) implies (not B) and that (not A) implies (B). This will establish that exactly one of (A) or (B) always occurs. Claim: (A) implies (not B). That is, if λ 1 < 1 and λ 2 < 1, then λ 1 < 1 or λ 2 < 1. Proof: By hypothesis, λ 1 < 1. So the claim holds. Claim: (not A) implies (B). That is, if λ 1 > 1 or λ 2 > 1, then λ 1 > 1 and λ 2 > 1. Proof: There are two cases to consider: one in which λ 1 > 1 and one in which λ 2 > 1. We must show that in both cases, λ 1 > 1 and λ 2 > 1. Case 1: Suppose λ 1 > 1. Then it follows that p - 2 > - p p 2 - 4 q . Seeking contra- diction, assume λ 2 < 1. This implies p - 2 < p p 2 - 4 q . So we have | p - 2 | < p p 2 - 4 q ⇐⇒ p 2 - 4 p + 4 < p 2 - 4 q ⇐⇒ p - q > 1 This is a contradiction of the given information that
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Problem_Set_5_Solutions - Problem Set 5 Solutions Economics...

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