Macroeconomics Exam Review 116

Macroeconomics Exam Review 116 - 3.49 Let be a hyperplane...

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Unformatted text preview: 3.49 Let be a hyperplane in . Then there exists a unique subspace such that = x + for some x (Exercise 1.153). There are two cases to consider. Case 1: x / . For every x , there exists unique x such x = x x + for some Define ( x ) = x . Then : . It is straightforward to show that is linear. Since = x + , x = 1 if and only if x . Therefore = { x : ( x ) = 1 } Case 2: x . In this case, choose some x 1 / . Again, for every x , there exists a unique x such x = x x 1 + for some and ( x ) = x is a linear functional on . Furthermore x implies = (Exercise 1.153) and therefore ( x ) = 0 if and only if x . Therefore = { x : ( x ) = 0 } Conversely, let be a nonzero linear functional in . Let...
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