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Unformatted text preview: Math 136 Term Test 2 Solutions 1. Short Answer Problems [2] a) What is the definition of a basis of a vector space V ? Solution: A basis of a vector space V is a linearly independent spanning set. [1] b) What is the definition of the dimension of a vector space V ? Solution: The dimension of a vector space is the number of vectors in any basis of V . [1] c) What is the definition of the columnspace of a matrix A ? Solution: The columnspace of a matrix is the span of the columns of A . [2] d) Let L : R n R m be a linear mapping. Prove that L ( ~ 0) = ~ 0. Solution: Since L is linear we have L ( t~x ) = tL ( ~x ) for any t R . Thus, since 0 ~x = ~ 0, we have L ( ~ 0) = L (0 ~x ) = 0 L ( ~x ) = ~ Alternate Proof: L ( ~ 0) = L ( ~x ~x ) = L ( ~x ) L ( ~x ) = ~ 0. [6] e) State the definition of a vector space V over R . Solution: A vector space V over R is a set with two operations of addition and scalar multiplication such that for any ~x,~ y,~ z V and any s,t R we have the following 10 properties: V1 ~x + ~ y V V2 ( ~x + ~ y ) + ~ z = ~x + ( ~ y + ~ z ) V3 There exists ~ V such that ~x + ~ 0 = ~x = ~ 0 + ~x for all ~x V V4 For each ~x V there is a ( ~x ) V such that ~x + ( ~x ) = ~ V5 ~x + ~ y = ~ y + ~x V6 s~x V V7 s ( t~x ) = ( st ) ~x V8 ( s + t ) ~x = s~x + t~x V9 s ( ~x + ~ y ) = s~x + s~ y V10 1 ~x = ~x 1 2 2. Let L : R 2 R 3 be defined by L ( x 1 ,x 2 ) = ( x 1 x 2 , x 1 + x 2 ,x 1 + 2 x 2 ). [2] a) Prove that L is linear. Solution: Let ~x = x 1 x 2 , ~ y = y 1 y 2 and t R . Then L ( t~x + ~ y ) = L ( tx 1 + y 1 ,tx 2 + y 2 ) = ( tx 1 + y 1 ) ( tx 2 + y 2 ) ( tx 1 + y 1 ) + ( tx 2 + y 2 ) ( tx 1 + y 1 ) + 2( tx 2 + y 2 ) = t x 1 x 2 x 1 + x 2 x 1 + 2 x 2 + y 1 y 2 y 1 + y 2 y 1 + 2 y 2 Hence, L is linear. [3] b) Find the standard matrix of L ....
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This note was uploaded on 11/25/2010 for the course MATH 136 taught by Professor All during the Spring '08 term at Waterloo.
 Spring '08
 All
 Linear Algebra, Algebra, Vector Space

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