Macroeconomics Exam Review 117

Macroeconomics Exam Review 117 - 3.51 By Exercise 3.49,...

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Unformatted text preview: 3.51 By Exercise 3.49, there exists a linear functional such that = { : ( ) = 0 } Since x / , ( x ) = 0. Without loss of generality, we can normalize so that ( x ) = 1. (If ( x ) = = 1, then the linear functional = 1 / c has ( x ) = 1 and kernel = .) To show that is unique, suppose that is another linear functional with kernel = and ( x ) = 1. For any x , there exists such that x = x + v with (Exercise 1.153). Since ( v ) = ( v ) = 0 and ( x ) = ( x ) = 1 ( x ) = ( x + v ) = ( ) = ( x ) = ( x + v ) = ( x ) 3.52 Assume = , = 0. Then ( ) = 0 ( ) = 0 Conversely, let = 1 (0) = 1 (0). If = , then = = . Otherwise, is a hyperplane containing . Choose some....
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This note was uploaded on 01/16/2012 for the course ECO 2024 taught by Professor Dr.dumond during the Fall '10 term at FSU.

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