chapter6 - Chapter 6 Solutions to 6.1 1. Let (x1, x2), (y1,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 6 Solutions to §6.1 1. Let (x 1 , x 2 ), (y 1 , y 2 ) R 2 and c R . T[(x 1 , x 2 )] = (x 1 + 2x 2 , 2x 1 – x 2 ) T[(x 1 , x 2 ) + (y 1 , y 2 )] = T[(x 1 + y 1 , x 2 + y 2 )] = [(x 1 + y 1 + 2x 2 + 2y 2 , 2x 1 + 2y 1 – x 2 – y 2 )] = (x 1 + 2x 2 + y 1 + 2y 2 , 2x 1 – x 2 + 2y 1 – y 2 ) = (x 1 + 2x 2 , 2x 1 – x 2 ) + (y 1 + 2y 2 , 2y 1 – y 2 ) = T[(x 1 , x 2 )] + T[(y 1 , y 2 )]. T[c(x 1 , x 2 )] = T[(cx 1 , cx 2 )] = (cx 1 + 2cx 2 , 2cx 1 – cx 2 ) = c(x 1 + 2x 2 , 2x 1 – x 2 ) = cT[(x 1 , x 2 )]. Thus, T is a linear transformation. 3. Let y 1 and y 2 be in C 2 (I), and let c be an arbitrary real number. Then, T(y 1 + y 2 ) = (y 1 + y 2 )´´ – 16(y 1 + y 2 ) = y 1 ´´ – 16y 1 + y 2 ´´ – 16y 2 = T(y 1 ) + T(y 2 ) T(cy 1 ) = (cy 1 )´´ – 16(cy 1 ) = c(y 1 ´´ – 16y 1 ) = cT(y 1 ). Consequently, T is a linear transformation. 5. Let f, g V, c R and T[f] = a b f(x)dx. T(f + g) = a b (f + g)(x)dx = a b [f(x) + g(x)]dx = a b f(x)dx + a b g(x)dx = T(f) + T(g). T(cf) = a b [cf(x)]dx = c a b f(x)dx = cT(f). Therefore, T is a linear transformation. 7. Let A and B be in M n ( R ), and let c be an arbitrary real number. Then T(A + B) = (A + B) + (A + B) T = A + A T + B + B T = T(A) + T(B). T(cA) = (cA) + (cA) T = c(A + A T ) = cT(A). Consequently, T is a linear transformation. 9. Let x = (x 1 , x 2 ), y = (y 1 , y 2 ) be in R 2 . Then 169
Background image of page 1

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

View Full DocumentRight Arrow Icon
170 Ch. 6 Linear Transformations and the Eigenvalue/Eigenvector Problem T( x + y ) = T((x 1 + y 1 , x 2 + y 2 ) = (x 1 + x 2 + y 1 + y 2 , 2), whereas T( x ) + T( y ) = (x 1 + x 2 , 2) + (y 1 + y 2 , 2) = (x 1 + x 2 + y 1 + y 2 , 4). We see that T( x + y ) T( x ) + T( y ), and therefore T is not a linear transformation. 11. If T(x 1 , x 2 ) = (3x 1 – 2x 2 , x 1 + 5x 2 ) then A = [T( e 1 ), T( e 2 )] = 3 –2 1 5 . 13. If T(x 1 , x 2 , x 3 ) = (x 1 – x 2 + x 3 , x 3 – x 1 ) then A = [T( e 1 ), T( e 2 ), T( e 3 )] = 1 –1 1 –1 0 1 . 15. T( x ) = A x = 1 3 –4 7 x 1 x 2 = x 1 + 3x 2 –4x 1 + 7x 2 , which we write as T(x 1 , x 2 ) = (x 1 + 3x 2 , –4x 1 + 7x 2 ). 17. T( x ) = 2 2 –3 4 –1 2 5 7 –8 x 1 x 2 x 3 = 2x 1 + 2x 2 – 3x 3 4x 1 – x 2 + 2x 3 5x 1 + 7x 2 – 8x 3 , which we write as T(x 1 , x 2 , x 3 ) = (2x 1 + 2x 2 – 3x 3 , 4x 1 – x 2 + 2x 3 , 5x 1 + 7x 2 – 8x 3 ). 19. a) If D = [ v 1 , v 2 ] = 1 1 1 –1 then det(D) = –2 0 so by Corollary 5.5.2 the vectors v 1 = (1, 1) and v 2 = (1, –1) are linearly independent. Since dim[ R 2 ] = 2, it follows from Theorem 5.6.4 that { v 1 , v 2 } is a basis for R 2 . b)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 40

chapter6 - Chapter 6 Solutions to 6.1 1. Let (x1, x2), (y1,...

This preview shows document pages 1 - 3. Sign up to view the full document.

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