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Chapter 5

# Chapter 5 - CHAPTER 5 Linear Transformations 5.1 Linear...

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CHAPTER 5 Linear Transformations 5.1 Linear Transformations Note: Many different arguments may be used to prove nonlinearity; our solutions to Problems 1–23 provide a sampling. ! Checking Linearity 1. T x y xy , ( ) = If u = u u 1 2 , a f , v = v v 1 2 , a f , T T u v u v u v u v T T u u v v u v u v + ( ) = + + = + + ( ) + ( ) = + 1 1 2 2 1 1 2 2 1 2 1 2 , . a f a fa f We see that T T T u v u v + ( ) ≠ ( ) + ( ) , so T is not a linear transformation. 2. T x y x y y , , ( ) = + ( ) 2 We can write this transformation in matrix form as T x y x y x y y , ( ) = L N M O Q P L N M O Q P = + L N M O Q P 1 1 0 2 2 . Hence, T is a linear transformation. 3. T x y xy y , , ( ) = ( ) 2 If we let u = u u 1 2 , a f , we have cT cT u u c u u u cu u cu u ( ) = = = 1 2 1 2 2 1 2 2 2 2 , , , a f a f a f 351

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352 CHAPTER 5 Linear Transformations and T c T cu cu c u u cu u ( ) = = 1 2 2 1 2 2 2 , , a f b g . Hence cT T c u u ( ) ≠ ( ) , so T is not a linear transformation. 4. T x y x x y , , ( ) = + ( ) 2, Note that T 0 0 0 2, 0 , , ( ) = ( ) . Linear transformations always map the zero vector into the zero vector (in their respective spaces), so T is not a linear transformation. 5. T x y x , , , ( ) = ( ) 0 0 We let u = u u 1 2 , a f , v = v v 1 2 , a f so T T u v u v u v u v T T u v u v + = + + = + = + = + b g b g b g b g b g b g b g 1 1 2 2 1 1 1 1 0 0 0 0 0 0 , , , , , , , and cT c u cu T c u u ( ) = = = ( ) 1 1 0 0 , , , 0 , 0 a f a f . Hence, T is a linear transformation from R 2 to R 3 . 6. T x y x y , , , , ( ) = ( ) 1 1 Because T does not map the zero vector 0 0 2 , R into the zero vector 0 0 0 0 4 , , , R , T is not a linear transformation. 7. T f f ( ) = ( ) 0 If f and g are continuous functions on 0 1 , , then T f g f g f g T f T g + ( ) = + ( )( ) = ( ) + ( ) = ( ) + ( ) 0 0 0 and T cf cf cf cT f ( ) = ( )( ) = ( ) = ( ) 0 0 . Hence, T is a linear transformation. 8. T f f ( ) = − If f and g are continuous functions on 0 1 , , then T f g f g f g T f T g + ( ) = − + ( ) = − = ( ) + ( ) and T cf cf c f cT f ( ) = − = ( ) = ( ) . Hence, T is a linear transformation.
SECTION 5.1 Linear Transformations 353 9. T f tf t ( ) = ′( ) If f and g are continuous functions on 0 1 , , then T f g t f t g t tf t tg t T f T g + ( ) = ( ) + ( ) ′ = ′( ) + ′( ) = ( ) + ( ) and T cf t cf t ctf t cT f ( ) = ( ) ( )′ = ′( ) = ( ) . Hence, T is a linear transformation. 10. T f f f f ( ) = ′′ + ′ + 2 3 If we are given that f and g are continuous functions that have two continuous derivatives, then T f g f g f g f g f f f g g g T f T g + ( ) = + ( )′′ + + ( )′ + + ( ) = ′′ + ′ + ( ) + ′′ + ′ + ( ) = ( ) + ( ) 2 3 2 3 2 3 and T cf cf cf cf c f f f cT f ( ) = ( )′′ + ( )′ + ( ) = ′′ + ′ + ( ) = ( ) 2 3 2 3 . Hence, T is a linear transformation. 11. T at bt c at b 2 2 + + = + b g If we introduce the two vectors p q = + + = + + a t b t c a t b t c 1 2 1 1 2 2 2 2 then T T a a t b b t c c a a t b b a t b a t b T T p q p q + ( ) = + + + + + = + + + = + + + = ( ) + ( ) 1 2 2 1 2 1 2 1 2 1 2 1 1 2 2 2 2 2 a f a f a f a f a f a f a f and T c T ca t cb t cc ca t cb c a t b cT p p ( ) = + + = + = + = ( ) 1 2 1 1 1 1 1 1 2 2 b g a f . Hence, the derivative transformation defined on P 2 is a linear transformation. 12. T at bt ct d a b 3 2 + + + = + b g If we introduce the two vectors p q = + + + = + + + a t b t c t d a t b t c t d 1 3 1 2 1 1 2 3 2 2 2 2

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354 CHAPTER 5 Linear Transformations then T T a a t b b t c c t d d a a b b a b a b T T T c T c a t b t c t d T ca t cb t cc t cd ca cb c a b cT p q p q p p + ( ) = + + + + + + + = + + + = + + + = ( ) + ( ) ( ) = + + + = + + + = + = + = ( ) 1 2 3 1 2 2 1 2 1 2 1 2 1 2 1 1 2 2 1 3 1 2 1 1 1 3 1 2 1 1 1 1 1 1 a f a f a f a f a f a f a f a f b g b g a f .
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Chapter 5 - CHAPTER 5 Linear Transformations 5.1 Linear...

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