InstructorsManualLeon7edPart2 - Chapter 4 SECTION 1 2. x1 =...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 Frst 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 - 2 3 5 2 = 7 18 63
Background image of page 1

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

View Full DocumentRight Arrow Icon
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 = + CB + AC + BC =( + AC + L ( A )+ L ( B ) Therefore L is a linear operator. (b) L ( αA + βB C 2 ( αA + 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 L ( I ) 10. If f,g C [0 , 1] then L ( αf + βg ± x 0 ( αf ( t ( t )) dt = α ± x 0 f ( t ) dt + β ± x 0 g ( t ) dt = αL ( f ( 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 ,...,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 )
Background image of page 2
Section 1 65 14. Let L be a linear transformation from R 1 to R 1 .I f 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 )) + 2 ( L 1 ( v 2 )) = αL ( v 1 ( v 2 ) Therefore L is a linear transformation. 17. (b) ker( L ) = Span( e 3 ), L ( R 3 ) = Span( e 1 , e 2 ) 18. (c) L ( S ) = Span((1 , 1 , 1) T ) 19. (b) If p ( x ax 2 + bx + c is in ker( L ), then L ( p )=( ax 2 + bx + c ) - (2 ax + b ax 2 +( b - 2 a ) x c - b ) must equal the zero polynomial z ( x )=0 x 2 +0 x + 0. Equating coefficients we see that a = b = c = 0 and hence ker( L { 0 } . The range of L is all of P 3 . To see this note that if p ( x ax 2 + bx + c is any vector in P 3 and we deFne q ( x ax 2 b +2 a ) x + c + b a then L ( q ( x )) = (
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 47

InstructorsManualLeon7edPart2 - Chapter 4 SECTION 1 2. x1 =...

This preview shows document pages 1 - 4. Sign up to view the full document.

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