2.1-2.3a

2.1-2.3a - 1 Math 110 Homework 4 Partial Solutions If you...

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1 Math 110 Homework 4 Partial Solutions If you have any questions about these solutions, or about any problem not solved, please ask via email or in office hours, etc. 2.1.13 Suppose that a 1 v 1 + ··· + a n v n = 0 for some a 1 , . . . , a n F . Applying T to both sides, we get T ( a 1 v 1 + ··· + a n v n ) = T (0) = 0. But 0 = T ( a 1 v 1 + ··· + a n v n ) = a 1 T ( v 1 ) + ··· + a n T ( v n ) = a 1 w 1 + ··· + a n w n . Since { w 1 , . . . , w n } is linearly independent, we have a 1 = ··· = a n = 0, and thus that { v 1 , . . . , v ) n } is linearly independent. 2.1.20 Suppose that V 1 is a subspace of V . First, 0 V 1 , so T (0) = 0 T ( V 1 ). Suppose that x, y T ( V 1 ). Then there are v, w V 1 such that T ( u ) = x and T ( v ) = y . Then T ( u + v ) = T ( u )+ T ( v ) = x + y T ( V 1 ). Suppose x T ( V 1 ) and a F . Then there is u V 1 such that T ( u ) = x . Then T ( au ) = aT ( u ) = ax T ( V 1 ). Thus T ( V 1 ) is a subspace of W . For the second statement, we first notice that T (0) = 0 W 1 , so 0 Z = { x V | T ( x ) W 1 } . Next, suppose that x, y Z . Then T ( x ) , T ( y ) W 1 . Thus T ( x + y ) = T ( x ) + T ( y ) W 1 since W 1 is a subspace. This gives that x + y Z . Suppose now that x Z and a F . Then T ( ax ) = aT ( x ) W 1 , since T ( x ) W 1 by assumption.
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2.1-2.3a - 1 Math 110 Homework 4 Partial Solutions If you...

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