sol4 - SOLUTIONS FOR HOMEWORK 4 Section 2.1 2. Note that T...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 one-to-one, since it has a non-trivial 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 one-to-one. 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 doesnt vanish at 0 cannot belong to R ( T )....
View Full Document

Page1 / 6

sol4 - SOLUTIONS FOR HOMEWORK 4 Section 2.1 2. Note that T...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online