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Unformatted text preview: Math 121: Linear Algebra and Applications Prof. Lydia Bieri Solution Set 4 Posted: Tue. Nov. 6, 2007 Written by: Luca Candelori Exercise 1 (2.1/15) . Linearity follows directly from the properties of integrals. Namely Z x ( f ( t ) + g ( t )) dt = Z x f ( t ) dt + Z x g ( t ) dt It is also 11 for suppose that R x f ( t ) dt = 0, the zero polynomial. From the Fundamental Theorem of Calculus: d dx Z x f ( t ) dt = f ( x ) = d dt (0) = 0 hence f ( x ) is the zero function, i.e. N ( T ) = 0 and the map is 11. Let now c be a constant in P ( R ). Suppose that there is a f ( x ) in P ( R ) such that R x f ( t ) dt = c . By applying derivatives to both sides we get that f ( x ) = 0, which is impossible, since the zero function integrates to 0, and not to c . Therefore there is no such f and the map T is not surjective. Exercise 2 (2.1/20) . Let u 1 ,u 2 T ( V 1 ) and c a non zero scalar in F . Then there exists v 1 ,v 2 V 1 such that T ( v 1 ) = u 1 and T ( v 2 ) = u 2 . Therefore cu 1 + u 2 = cT ( v 1 ) + T ( v 2 ) = T ( cv 1 + v 2 ) Since V 1 is a subspace, cv 1 + v 2 V 1 and therefore cu 1 + u 2 T ( V 1 ). So T ( V 1 ) is a subspace. Let now S = { x V  T ( x ) W 1 } . We want to show this is a subspace of V . Take x 1 ,x 2 S and c a nonzero scalar in F . Then T ( cx 1 + x 2 ) = cT ( x 1 ) + T ( x 2 ) W 1 since T ( x 1 ) ,T ( x 2 ) W 1 and W 1 is a subspace of W . Therefore cx 1 + x 2 S ....
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This note was uploaded on 10/23/2010 for the course MATH Math 121 taught by Professor Bieri during the Fall '07 term at Harvard.
 Fall '07
 bieri
 Math, Linear Algebra, Algebra, Integrals

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