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Chapter 4 ( B )

Chapter 4 ( B ). General Vector Spaces(I) (4.3 ~ 4.5) Gareth Williams J & B, 滄海書局. Part B. 4.3 General Vector Spaces 4.4 Subspace 4.5 Linear Combinations of Vectors. 4.3 Vector Spaces. Definition

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Chapter 4 ( B )

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  1. Chapter 4(B) General Vector Spaces(I) (4.3 ~ 4.5) Gareth Williams J & B, 滄海書局

  2. Part B • 4.3 General Vector Spaces • 4.4 Subspace • 4.5 Linear Combinations of Vectors

  3. 4.3 Vector Spaces • Definition • A vector space is a set V of elements called vectors, having operations of addition and scalar multiplication defined on it that satisfy the following conditions (u, v, and w are arbitrary elements of V, and c and d are scalars.) • Closure Axioms • The sum u + v exists and is an element of V. (V is closed under addition.) • cu is an element of V. (V is closed under scalar multiplication.)

  4. Addition Axioms 3. u + v = v + u (commutative property) 4. u + (v + w) = (u + v) + w (associative property) 5. There exists an element of V, called the zero vector, denoted 0, such that u + 0 = u. 6. For every element u of V there exists an element called the negative of u, denoted -u, such that u + (-u) = 0. Scalar Multiplication Axioms 7.c(u + v) = cu + cv 8. (c + d)u =cu + du 9. c(du) = (cd)u 10.1u = u

  5. Vector Spaces of Matrices Use vector notation for the elements of M22. Let Be two arbitrary 2  2 matrices. Closure Axioms:

  6. Addition Axioms:

  7. Scalar Multiplication Axioms:

  8. Note Mmn, the set of m n matrices, is a vector space.

  9. Vector Spaces of Functions Closure Axioms: Axiom 1: f + g is defined by (f + g)(x) = f(x) + g(x). f + g is thus a functions with domain the set of real numbers. f + g is an element of V. Thus V is closed under addition. Axiom 2: cf is defined by (cf)(x) = c[f(x)]. cf is thus a functions with domain the set of real numbers. cf is an element of V. Thus V is closed under scalar multiplication.

  10. Vector Spaces of Functions Addition Axioms:

  11. Axiom 6: The value of the function [f + (-f )] is the same as the value of 0 at every x. Thus [f + (-f )] = 0. Therefore -f is the negative of f. Vector Spaces of Functions Axiom 5: Let 0 be the function such that 0(x) = 0 for every real number x. 0 is called the zero function. We get (f + 0)(x) = f(x) + 0(x) = f(x) + 0 = f(x) for every real number x. The value of the function f + 0 is the same as the value of f at every x. Thus f + 0 = f. 0 is the zero vector.

  12. Vector Spaces of Functions Scalar Multiplication Axioms:

  13. The complex Vector Space Cn

  14. Theorem 4.5 Let V be a vector space, v a vector in V, 0 the zero vector of V, c a scalar, and 0 the zero scalar. Then (a) 0v = 0 (b)c0 = 0 (c) (-1)v = -v (d) If cv = 0, then either c = 0 or v = 0.

  15. Theorem 4.5 Proof (a) 0v + 0v = (0 + 0)v = 0v (0v + 0v) + (-0v) = 0v + (-0v) 0v + [(0v + (-0v)] = 0, 0v + 0 = 0, 0v = 0 (b) c0=c0+0=c0+(c0+-(c0)) =(c0+c0)+-(c0) =c(0+0)+-(c0) =c0+-(c0)=0 (c) (-1)v + v = (-1)v + 1v = [(-1) + 1]v = 0v = 0

  16. 4.4 Subspaces Definition Let V be a vector space and U be a nonempty subset of V. U is said to be a subspace of V if it is closed under addition and under scalar multiplication. Figure 4.16

  17. Example 1 Let U be the subset of R3 consisting of all vectors of the form (a, 0, 0) (with zeros as second and third components). Show that U is a subspace of R3. Solution

  18. Example 2 Let W be the set of vectors of the form (a, a2, b). Show that W is not a subspace of R3. Solution

  19. Example 3 Consider R3 with vectors written in column form. Let U be the subset of vectors of the from .Is U a subspace of R3. Solution

  20. Example 4 Prove that the set 2  2 diagonal matrices is a subspace of the vector space M22 of 2  2 matrices. Solution

  21. Example 5 Let Pn denoted the set of real polynomial functions of degree  n. Prove that Pn is a vector space if addition and scalar multiplication are defined on polynomials in a pointwise manner. Solution

  22. Proof Let u be an arbitrary vector in U and 0 be the vector of V. Let 0 be the zero scalar. By Theorem 4.5(a) we know that 0u = 0. Since U is closed under scalar multiplication, this means that 0 is in U. Theorem 4.6 Let U be a subspace of a vector space V. U contains the zero vector of V.

  23. Example 6 Let W be the set of vectors of the form (a, a, a+2). Show that W is not a subspace of R3. Solution

  24. 4.5 Linear Combinations of Vectors (a, a, b) = a(1, 1, 0) + b(0, 0, 1) (2, 2, 3) = 2(1, 1,0) +3(0, 0, 1) (-1, -1, 7) = -1 (1, 1,0) + 7(0, 0, 1)

  25. Definition Let v1, v2, …, vm be vectors in a vector space V. We say that v, a vector V, is a linear combination of v1, v2, …, vm if there exist scalars c1, c2, …, cm such that v can be written v = c1v1 + c2v2 + … + cmvm

  26. Example 1 The vector (5, 4, 2) is a linear combination of the vector (1, 2, 0), (3, 1, 4) and (1, 0, 3) since it can be written (5, 4, 2) = (1, 2, 0) + 2(3, 1, 4) – 2(1, 0, 3)

  27. Example 2 Determine whether or not the vector (-1, 1, 5) is a linear combination of the vectors (1, 2, 3), (0, 1, 4), and (2, 3, 8). Solution

  28. Example 3 Express the vector (4, 5, 5) as a linear combination of the vectors (1, 2, 3), (-1, 1, 4), and (3, 3, 2). Solution

  29. Example 4 Show that the vector (3, -4, -6) cannot be expressed as a linear combination of the vectors (1, 2, 3), (-1, -1, -2), and (1, 4, 5). Solution

  30. Example 5 Is the vector a linear combination of , , and ? Solution

  31. Determine whether the matrix is a linear combination of the matrices in the vector space M22 of 2  2 matrices. Example 6 Solution

  32. Example 7 Determine whether the function f(x)=x2+10x-7 is a linear combination of the function g(x)=x2+3x-1 and h(x)=2x2-x+4. Solution

  33. Definition The vectors v1, v2, …, vm are said to span a vector space if every vector in the space can be expressed as a linear combination of these vectors.

  34. Example 8 Show that the vectors (1, 2, 0), (0, 1, -1), and (1, 1, 2) span R3. Solution

  35. Example 8

  36. Example 9 Show that the following matrices span the vector space M22 of 2  2 matrices. Solution

  37. Proof Let u1 = a1v1 + … + amvm and u2 = b1v1 + … + bmvm be arbitrary elements of U. Then u1 + u2 = (a1v1 + … + amvm) + (b1v1 + … + bmvm) = (a1 + b1) v1 + … + (am + bm) vm u1 + u2 is a linear combination of v1, …, vm . Thus u1 + u2 is in U. U is closed under vector addition. Theorem 4.7 Let v1, …, vm be vectors in a vector space V. Let U be the set consisting of all linear combinations of v1, …, vm . Then U is a subspace of V spanned the vectors v1, …, vm . U is said to be the vector space generated by v1, …, vm .

  38. Theorem 4.7 Let c be an arbitrary scalar. Then cu1 = c(a1v1 + … + amvm) = ca1v1 + … + camvm) cu1 is a linear combination of v1, …, vm . Therefore cu1 is in U. U is closed under scalar multiplication. Thus U is a subspace of V. By the definition of U, every vector in U can be written as a linear combination of v1, …, vm . Thus v1, …, vm span U.

  39. We can visualize U. U is made up of all vectors in the plane defined by the vectors (-1, 5, 3) and (2, -3, 4). Example 10 Consider the vector space R3. The vectors (-1, 5, 3) and (2, -3, 4) are in R3. Let U be the subset of R3 consisting of all vectors of the form c1(-1, 5, 3) + c2(2, -3, 4) Then U is a subspace of R3 spanned by (-1, 5, 3) and (2, -3, 4). The following are examples of some of the vectors in U, obtained by given c1 and c2 various values.

  40. Figure 4.17

  41. Figure 4.18 We can generalize this result. Let v1 and v2 be vectors in the space R3. The subspace U generated by v1 and v2 is the set of all vectors of the form c1v1 + c2v2. If v1 and v2 are not colinear, then U is the plane defined by v1 and v2 .

  42. Example 11 Let v1 and v2 span a subspace U of a vector V. Let k1 and k2 be nonzero scalars. Shoe that k1v1 and k2v2 also span U. Solution

  43. Example 11 Figure 4.19

  44. Example 12 Let U be the subspace of R3 generated by the vectors (1, 2, 0) and (-3, 1, 2). Let V be the subspace of R3 generated by the vectors (-1, 5,2) and (4, 1, -2). Show that U = V. Solution

  45. Example 12 Figure 4.20

  46. Example 13 Let U be the vector space generated by the functions f(x) = x + 1 and g(x) = 2x2 – 2x + 3. Show that the function h(x) = 6x2 – 10x + 5 lies in U. Solution

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