4 4 coordinate systems in general people are more
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4.4 Coordinate Systems In general, people are more comfortable - PDF document

4.4 Coordinate Systems In general, people are more comfortable working with the vector space R n and its subspaces than with other types of vectors spaces and subspaces. The goal here is to impose coordinate systems on vector spaces, even if they


  1. 4.4 Coordinate Systems In general, people are more comfortable working with the vector space R n and its subspaces than with other types of vectors spaces and subspaces. The goal here is to impose coordinate systems on vector spaces, even if they are not in R n . THEOREM 7 The Unique Representation Theorem Let β =  b 1 , … , b n  be a basis for a vector space V . Then for each x in V , there exists a unique set of scalars c 1 , … , c n such that x = c 1 b 1 + ⋯ + c n b n . DEFINITION Suppose β =  b 1 , … , b n  is a basis for a vector space V and x is in V . The coordinates of x relative to the basis β (or the β − coordinates of x ) are the weights c 1 , … , c n such that x = c 1 b 1 + ⋯ + c n b n . In this case, the vector in R n c 1  x  β = ⋮ c n is called the coordinate vector of x ( relative to β ), or the β − coordinate vector of x . 1

  2. 3 EXAMPLE: Let β =  b 1 , b 2  where b 1 = and 1 0 1 b 2 = and let E =  e 1 , e 2  where e 1 = and 1 0 0 e 2 = . 1 Solution: 2 If  x  β = , then 3 3 0 x = ____ . + ____ = 1 1 6 If  x  E = , then 5 1 0 x = ____ + ____ . = 0 1 2

  3. x 2 7 6 5 4 3 2 1 x 1 1 2 3 4 5 6 7 Standard graph paper β − graph paper 3

  4. From the last example, 6 3 0 2 . = 5 1 1 3 For a basis β =  b 1 , … , b n  , let c 1 c 2 b 1 b 2 ⋯ b n and  x  β = P β = ⋮ c n Then x = P β  x  β . We call P β the change-of-coordinates matrix from β to the standard basis in R n . Then − 1 x  x  β = P β − 1 is a change-of-coordinates matrix from the and therefore P β standard basis in R n to the basis β . 4

  5. 3 0 EXAMPLE: Let b 1 = , b 2 = , β =  b 1 , b 2  and 1 1 6 x = . Find the change-of-coordinates matrix P β from β to 8 the standard basis in R 2 and change-of-coordinates matrix P β − 1 from the standard basis in R 2 to β . Solution b 1 b 2 and so P β = = − 1 1 0 3 0 3 − 1 = P β = − 1 1 1 1 3 6 2 − 1 to find  x  β = (b) If x = , then use P β . 8 6 1 0 6 3 − 1 x = Solution:  x  β = P β = − 1 1 8 3 5

  6. Coordinate mappings allow us to introduce coordinate systems for unfamiliar vector spaces. Standard basis for P 2 :  p 1 , p 2 , p 3  =  1, t , t 2  Polynomials in P 2 behave like vectors in R 3 . Since a + bt + ct 2 = ____ p 1 + ____ p 2 + ____ p 3 , a  a + bt + ct 2  β = b c We say that the vector space R 3 is isomorphic to P 2 . 6

  7. EXAMPLE: Parallel Worlds of R 3 and P 2 . Vector Space R 3 Vector Space P 2 a Vector Form: a + bt + bt 2 Vector Form: b c Vector Addition Example Vector Addition Example − 1 2 1  − 1 + 2 t − 3 t 2  +  2 + 3 t + 5 t 2  2 3 5 + = = 1 + 5 t + 2 t 2 − 3 5 2 Informally, we say that vector space V is isomorphic to W if every vector space calculation in V is accurately reproduced in W , and vice versa . Assume β is a basis set for vector space V . Exercise 25 (page 254) shows that a set  u 1 , u 2 , … , u p  in V is linearly independent if and only if  u 1  β ,  u 2  β , … ,  u p  β is linearly independent in R n . 7

  8. EXAMPLE: Use coordinate vectors to determine if  p 1 , p 2 , p 3  is a linearly independent set, where p 1 = 1 − t , p 2 = 2 − t + t 2 , and p 3 = 2 t + 3 t 2 . Solution: The standard basis set for P 2 is β =  1, t , t 2  . So  p 1  β = ,  p 2  β = ,  p 3  β = Then 1 2 0 1 2 0 − 1 − 1 2 0 1 2  ⋯  0 1 3 0 0 1 By the IMT,  p 1  β ,  p 2  β ,  p 3  β is linearly ____________________ and therefore  p 1 , p 2 , p 3  is linearly ____________________. Coordinate vectors also allow us to associate vector spaces with subspaces of other vectors spaces. 8

  9. 3 EXAMPLE Let β =  b 1 , b 2  where b 1 = and 3 1 0 9 b 2 = and let H = span  b 1 , b 2  . Find  x  β , if x = . 1 13 3 15 Solution: (a) Find c 1 and c 2 such that 3 0 9 c 1 + c 2 3 1 13 = 1 3 15 Corresponding augmented matrix: 3 0 9 1 0 3 3 1 13 0 1 4 ∽ 1 3 15 0 0 0 Therefore c 1 = ____ and c 2 = _____ and so  x  β = . 9

  10. x 3 x 2 x 1 9 3 in R 3 is associated with the vector in R 2 13 4 15 H is isomorphic to R 2 10

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