Mid1_soln

Mid1_soln - MA527 MID-TERM EXAM#1 Solution of Problem#1...

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Unformatted text preview: MA527: MID-TERM EXAM #1 (SEPTEMBER 23, 2008) Solution of Problem #1. First, check the “zero” element, that is, let a function f ( x ) = 0 , ≤ x ≤ 1. In this case, f (0) = 0. Therefore, if a 6 = 0 in the definition of V , then f ( x ) / ∈ V . So, V is not a vector space if a 6 = 0. If a = 0, then the function f ( x ) = 0 belongs to the space because it is continuous and f (0) = 0 = a . Let g ( x ) ∈ V , that is g ( x ) is continuous and g (0) = 0. Then- g ( x ) is also continuous and- g (0) = 0, so- g ( x ) ∈ V . And αg ( x ) is continous and αg (0) = α · 0 = 0, and belongs to V . Let another function h ( x ) be continuous with h (0) = 0, that is, h ( x ) ∈ V . Then g ( x ) + h ( x ) is still continuous and g (0) + h (0) = 0, and g ( x ) + h ( x ) ∈ V . So, V is a vector space if a = 0. Solution to Problem #2. A = 2 | 1 3 | 1 4 | 1 1 | 1 → 1 | 1 2 | 1 3 | 1 4 | 1 This is done by simply exchanging rows. After dividing the 2nd row by 2, 3rd row by 3, and 4th row by 4, we obtain 1 | 1 1 | 1 2...
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Mid1_soln - MA527 MID-TERM EXAM#1 Solution of Problem#1...

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