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. Tx y xy , () = If u = uu 12 af , v = vv , T T uvuv uvuv TTu u v v uv + =+ + = + + () + () 1122 112 2 , . a f a fa f We see that TT T u v + ≠ ()+ () , so T is not a linear transformation. 2. x y y ,, 2 We can write this transformation in matrix form as x y y = L N M O Q P L N M O Q P = + L N M O Q P 11 02 2 . Hence, T is a linear transformation. 3. xy y = 2 If we let u = a f , we have cT cT u u c u u u cu u cu u == = 1 2 2 1 2 2 22 , a f a f 351
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352 CHAPTER 5 Linear Transformations and Tc Tcu cu cuu cu u () == 12 2 2 2 ,, a f bg . Hence cT T c uu () ≠() , so T is not a linear transformation. 4. Tx y x x y =+ 2, Note that T 00 02 ,0 = . Linear transformations always map the zero vector into the zero vector (in their respective spaces), so T is not a linear transformation. 5. x , = We let u = , af , v = vv so TT u v u v u v u vT T uv u v += + + = + = + = + b g b g b g b g b g b g 11 2 2 1 1 , , , , , and cT c u cu T c === , 0 a f . Hence, T is a linear transformation from R 2 to R 3 . 6. x y , , = Because T does not map the zero vector 2 R into the zero vector 0000 4 ,,, R , T is not a linear transformation. 7. Tf f () = () 0 If f and g are continuous functions on 01 , , then Tf g f g f g Tg + ( ) = ()+ () =()+ () 0 and f c f c f cT f = ( ) = () = () . Hence, T is a linear transformation. 8. f ()=− If f and g are continuous functions on , , then f g Tf + =− + =− − =()+ () and f c f c f ()=− = − = () . Hence, T is a linear transformation.
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SECTION 5.1 Linear Transformations 353 9. Tf t f t () = () If f and g are continuous functions on 01 , , then Tf g tft gt t f t t gt Tf Tg + = ()+ () ′ = ()+ () =() + () and Tc f tc c t cT f = = () = () . Hence, T is a linear transformation. 10. f f f ′′ + ′ + 23 If we are given that f and g are continuous functions that have two continuous derivatives, then Tf g f g f f f g g g Tf + =+ ′′ ++ = ′′ + ′ + + ′′ + ′ + =()+ () 2 32 3 and f c f c f c f cf f f = + + = ′′ + ′ + = () 2 3 . Hence, T is a linear transformation. 11. Ta t b t c a t b 2 2 ++= + bg If we introduce the two vectors p q + + at bt c 1 2 11 2 2 22 then TT a a t b b t c c a a t b b at b T T pq + = + = + = () + () 12 2 2 a f af a f a f and T c T ca t cb t cc ca t cb c a t b cT pp + = + = + = 1 2 1 1 . Hence, the derivative transformation defined on P 2 is a linear transformation. 12. t b t c t d a b + = + If we introduce the two vectors p q =++ + + bt ct d 1 3 1 2 2 3 2 2
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354 CHAPTER 5 Linear Transformations then TT a a t b b t c c t d d a a b b ab ab T T T c T c a t b t c t d T ca t cb t cc t cd ca cb ca b cT pq p p + () = + ++ =+++ =+++= ()+ () =+ + + = + + + = + = 12 3 2 1 2 12 12 11 2 2 1 3 1 2 1 3 1 2 111 1 a f af a f a f a f bg b g . Hence, the derivative transformation defined on P 3 is a linear transformation. 13. T AA () = T . If we introduce two 22 × matrices B and C , we have T Tk k k kT BC BC B C B C BB B B + =+= () +() = == T T T . Hence, the transformation defined on M 22 is a linear transformation. 14. T ab cd L N M O Q P = Letting A = L N M O Q P be an arbitrary vector, we show the homogeneous property = () fails because T ka kb kc kd ka kb kc kd kad kcb k A A = L N M O Q P = () ≠ () 222 2 det when k 1. Hence, T is not a linear transformation.
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Chapter 5 - CHAPTER 5 Linear Transformations 5.1 Linear...

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