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Unformatted text preview: SOLUTIONS FOR HOMEWORK 4 Section 2.1 2. Note that T ( a 1 ,a 2 ,a 3 ) = 0 iff a 1 = a 2 and a 3 = 0. Thus, N ( T ) = { ( a,a, 0) : a ∈ R } = span( { (1 , 1 , 0) } ) . Therefore, nullity( T ) = dim N ( T ) = 1. Furthermore, R ( T ) = R 2 . Indeed, we have to show that, for every ( b 1 ,b 2 ) ∈ R 2 , there exists ( a 1 ,a 2 ,a 3 ) ∈ R 3 s.t. T ( a 1 ,a 2 ,a 3 ) = ( a 1 − a 2 , 2 a 3 ) = ( b 1 ,b 2 ) . ( b 1 , ,b 2 / 2) will do. Thus, rank ( T ) = dim R ( T ) = 2. By Theorem 2.4, T is not onetoone, since it has a nontrivial null space. For instance, (1 , 1 , 0) ∈ N ( T ) \{ } . On the other hand, T is onto ( R ( T ) = R 2 ). 9. (b) We have to show that either (i) T does not preserve multiplication by a scalar, or (ii) T does not preserve addition. Actually, T does not preserve either. To be more precise: (i) Suppose x = (1 , 0), and c = 2. Then T ( x ) = (1 , 1), while T (2 x ) = (2 , 4). In particular, 2 T ( x ) negationslash = T (2 x ). (ii) Let x = y = (1 , 0). Then T ( x ) + T ( y ) = (2 , 2) negationslash = (2 , 4) = T ( x + y ). 10. To describe a linear transformation T : R 2 → W ( W is a vector space), it suffices to discover u 1 = T ( e 1 ) and u 2 = T ( e 2 ), where e 1 = (1 , 0) and e 2 = (0 , 1). Indeed, then we have T ( a 1 ,a 2 ) = T ( a 1 e 1 + a 2 e 2 ) = a 1 T ( e 1 ) + a 2 T ( e 2 ) = a 1 u 1 + a 2 u 2 . We have u 1 = T ( e 1 ) = (1 , 4). Moreover, T (1 , 1) = T ( e 1 + e 2 ) = u 1 + u 2 = (2 , 5) , hence u 2 = (2 , 5) − (1 , 4) = (1 , 1). Thus, T ( a 1 ,a 2 ) = a 1 (1 , 4) + a 2 (1 , 1) = ( a 1 + a 2 , 4 a 1 + a 2 ). In particular, T (2 , 3) = (5 , 11). T is onetoone. Indeed, by Theorem 2.4, we have to show that N ( T ) = { } . But ( a 1 ,a 2 ) ∈ N ( T ) iff a 1 + a 2 = 4 a 1 + a 2 = 0. Solving this system of linear equations, we see that a 1 = a 2 = 0. 13. We have to show that a 1 v 1 + ... + a n v n = 0 implies a 1 = ... = a n = 0. If the first equality holds, then 0 = T (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 . 1 2 SOLUTIONS FOR HOMEWORK 4 However, the set { w 1 ,...,w n } is linearly independent, hence a 1 = ... = a n = 0. 15. Consider f ( x ) = a + a 1 x + ... + a n x n . Let g = T ( f ). It is easy to see that g ( x ) = a x + a 1 x 2 / 2 + a 2 x 3 / 3 + a n x n +1 / ( n + 1). Thus, T is not onto. Indeed, in the above notation, g (0) = 0. Therefore, any polynomial h which doesn’t vanish at 0 cannot belong to R ( T )....
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.
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