This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: HOMEWORK 3  SOLUTIONS 1. (GRADED) (a) Pick u U \ { } such that T ( u ) 6 = 0. If v / U , then u + v / U otherwise v = ( u + v )+( 1 u ) U since U is closed under addition and scalar multiplication. Hence T ( u + v ) = 0, but T ( u ) + T ( v ) = T ( u ) + 0 = T ( u ) 6 = 0. (b) Clearly { e 1 ,e 2 } is a basis for U and { e 1 ,e 2 ,e 3 } is a basis of R 3 . Let T be the unique linear transformation such that T ( e i ) = e i for i = 1 , 2 and T ( e 3 ) = 0. Such T exists by the theorem on Extending by Linearity. Then for u U , say u = c 1 e 1 + c 2 e 2 , we have that: T ( u ) = c 1 T ( e 1 ) + c 2 T ( e 2 ) = c 1 e 1 + c 2 e 2 = u Hence T U = T as desired. (c) To show the map is linear, pick a R and T L ( R 3 , R ). First we need to show that ( aT ) = a ( T ). That is, ( aT ) U = a ( T U ). We show two functions are equal if they do the same thing to an arbitrary input. So pick u U arbitrary. Then: ( aT ) U ( u ) = ( aT )( u ) = a ( T ( u )) = a ( T U ( u )) = ( aT U )( u ) as desired. We also need to show that for T 1 ,T 2 L ( R 3 , R ), ( T 2 + T 2 ) = ( T 1 ) + ( T 2 ). That is, we need to show that ( T 1 + T 2 ) U = T 1 U + T 2 U . So again, we pick an arbitrary u U and compute: ( T 1 + T 2 ) U ( u ) = ( T 1 + T 2 )( u ) = T 1 ( u ) + T 2 ( u ) = T 1 U ( u ) + T 2 U ( u ) = ( T 1 U + T 2 U )( u ) as desired....
View Full
Document
 Fall '08
 GUREVITCH
 Math, Addition, Multiplication, Scalar

Click to edit the document details