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HW5_sol - Chapter 4 SECTION 1 2 x1 = r cos x2 = r sin where...

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Chapter 4 SECTION 1 2. x 1 = r cos θ , x 2 = r sin θ where r = ( x 2 1 + x 2 2 ) 1 / 2 and θ is the angle between x and e 1 . L ( x ) = ( r cos θ cos α - r sin θ sin α, r cos θ sin α + r sin θ cos α ) T = ( r cos( θ + α ) , r sin( θ + α )) T The linear transformation L has the effect of rotating a vector by an α in the counterclockwise direction. 3. If α = 1 then L ( α x ) = α x + a = α x + α a = αL ( x ) The addition property also fails L ( x + y ) = x + y + a L ( x ) + L ( y ) = x + y + 2 a 4. Let u 1 = 1 2 , u 2 = 1 - 1 , x = 7 5 To determine L ( x ) we must first express x as a linear combination x = c 1 u 1 + c 2 u 2 To do this we must solve the system U c = x for c . The solution is c = (4 , 3) T and it follows that L ( x ) = L (4 u 1 + 3 u 2 ) = 4 L ( u 1 ) + 3 L ( u 2 ) = 4 - 2 3 + 3 5 2 = 7 18 63
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64 Chapter 4 8. (a) L ( αA ) = C ( αA ) + ( αA ) C = α ( CA + AC ) = αL ( A ) and L ( A + B ) = C ( A + B ) + ( A + B ) C = CA + CB + AC + BC = ( CA + AC ) + ( CB + BC ) = L ( A ) + L ( B ) Therefore L is a linear operator. (b) L ( αA + βB ) = C 2 ( αA + βB ) = αC 2 A + βC 2 B = αL ( A ) + βL ( B ) Therefore L is a linear operator. (c) If C = O then L is not a linear operator. For example, L (2 I ) = (2 I ) 2 C = 4 C = 2 C = 2 L ( I ) 10. If f, g C [0 , 1] then L ( αf + βg ) = x 0 ( αf ( t ) + βg ( t )) dt = α x 0 f ( t ) dt + β x 0 g ( t ) dt = αL ( f ) + βL ( g ) Thus L is a linear transformation from C [0 , 1] to C [0 , 1]. 12. If L is a linear operator from V into W use mathematical induction to prove L ( α 1 v 1 + α 2 v 2 + · · · + α n v n ) = α 1 L ( v 1 ) + α 2 L ( v 2 ) + · · · + α n L ( v n ) . Proof: In the case n = 1 L ( α 1 v 1 ) = α 1 L ( v 1 ) Let us assume the result is true for any linear combination of k vectors and apply L to a linear combination of k + 1 vectors. L ( α 1 v 1 + · · · + α k v k + α k +1 v k +1 ) = L ([ α 1 v 1 + · · · + α k v k ] + [ α k +1 v k +1 ]) = L ( α 1 v 1 + · · · + α k v k ) + L ( α k +1 v k +1 ) = α 1 L ( v 1 ) + · · · + α k L ( v k ) + α k +1 L ( v k +1 ) The result follows then by mathematical induction. 13. If v is any element of V then v = α 1 v 1 + α 2 v 2 + · · · + α n v n Since L 1 ( v i ) = L 2 ( v i ) for i = 1 , . . ., n , it follows that L 1 ( v ) = α 1 L 1 ( v 1 ) + α 2 L 1 ( v 2 ) + · · · + α n L 1 ( v n ) = α 1 L 2 ( v 1 ) + α 2 L 2 ( v 2 ) + · · · + α n L 2 ( v n ) = L 2 ( α 1 v 1 + · · · + α n v n ) = L 2 ( v )
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Section 1 65 14. Let L be a linear transformation from R 1 to R 1 . If L ( 1 ) = a then L ( x ) = L ( x 1 ) = xL ( 1 ) = x a = a x 15. The proof is by induction on n . In the case n = 1, L 1 is a linear operator since L 1 = L . We will show that if L m is a linear operator on V then L m +1 is also a linear operator on V . This follows since L m +1 ( α v ) = L ( L m ( α v )) = L ( αL m ( v )) = αL ( L m ( v )) = αL m +1 ( v ) and L m +1 ( v 1 + v 2 ) = L ( L m ( v 1 + v 2 )) = L ( L m ( v 1 ) + L m ( v 2 )) = L ( L m ( v 1 )) + L ( L m ( v 2 )) = L m +1 ( v 1 ) + L m +1 ( v 2 ) 16. If v 1 , v 2 V , then L ( α v 1 + β v 2 ) = L 2 ( L 1 ( α v 1 + β v 2 )) = L 2 ( αL 1 ( v ) + βL 1 ( v 2 )) = αL 2 ( L 1 ( v 1 )) + βL 2 ( L 1 ( v 2 )) = αL ( v 1 ) + βL ( v 2 ) Therefore L is a linear transformation.
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