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Macroeconomics Exam Review 181

# Macroeconomics Exam Review 181 - ± z = 1 x,³ = x 1 ³ ≥...

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3.240 We have already shown (Exercise 3.239) that the alternatives I and II are mutu- ally incompatible. If Gordan’s system II 𝐴 𝑇 y = 0 has a semipositive solution y 0 , then we can normalize y such that 1y = 1 and the system 𝐴 𝑇 y = 0 1y = 1 has a nonnegative solution. Conversely, if Gordan’s system II has no solution , the system 𝐵 y = c where 𝐵 = ( 𝐴 𝑇 1 ) and c = ( 0 , 1) = (0 , 0 , . . . , 0 , 1), 0 ∈ ℜ 𝑚 , is the ( 𝑚 + 1)st unit vector has no solution y 0. By the Farkas lemma, there exists z ∈ ℜ 𝑛 +1 such that 𝐵 z 0 cz < 0 Decompose z into z = ( x , 𝑥 ) with x ∈ ℜ 𝑛 . The second inequality implies that 𝑥 < 0 since cz = ( 0 , 1) ( x , 𝑥 ) = 𝑥 < 0 Since 𝐵 = ( 𝐴, 1 ), the first inequality implies that
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Unformatted text preview: ± z = ( ?, 1 )( x ,³ ) = ? x + 1 ³ ≥ or ? x ≥ − 1 ³ > x solves Gordan’s system I. 3.241 Let a 1 , a 2 ,..., a ± be a basis for ´ . Let ? = ( a 1 , a 2 ,..., a ± ) be the matrix whose columns are a ³ . To say that ´ contains no positive vector means that the system ? x > has no solution. By Gordan’s theorem, there exists some y ≩ such that ? ? y = that is a ³ y = ya ³ = 0 , µ = 1 , 2 ,...,² so that y ∈ ´ ⊥ . 190...
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