Chapter 4: Vector Spaces - Part 3/Slides By Pearson
- 2. Basis Defn - A set of vectors S = { v1, v2, , vk } in a vector space V is called a basis for V if and only if S is a) linearly independent and a) S spans V.
- 3. Standard (natural) basis for R2 is R2=Span Also R2=Span Therefore, is another basis for R2 Example of a basis for R2 1 2 1 0 S , , 0 1 e e 𝐞 , 𝐞 2 1 , 1 3 2 1 , 1 3
- 4. Standard (natural) basis for R3 is R3=Span 1 2 3 1 0 0 S 0 , 1 , 0 , , 0 0 1 e e e Example of a basis for R3 1 2 3, ,e e e
- 5. The following three-dimensional vectors are linearly independent. Any vector in R3 can be expressed as a linear combination of v1, v2 and v3. v1, v2 and v3 form a basis of R3. Span {v1, v2 , v3} = R3. 1 2 3 1 1 1 2 , 0 , 1 1 2 0 v v v Example of a basis for R3
- 6. The standard (natural) basis for Rn is denoted by e1, e2, , en where 0 0 th row1 0 0 i ie Example of a basis for Rn
- 7. Theorem - If S = { v1, v2, , vn } is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of the vectors in S Theorem - If S = { v1, v2, , vn } is a set of vectors spanning a vector space V, then S contains a basis T for V Basis
- 8. Theorem - If S = { v1, v2, , vn } is a basis for a vector space V and T = { w1, w2, , wr } is a linearly independent set of vectors in V, then r ≤ n. Corollary - If S = { v1, v2, , vn } and T = { w1, w2, , wm } are bases for a vector space V, then n = m, i.e. every basis for V contains the same number of vectors. Basis
- 9. Defn - The dimension of a nonzero vector space V is the number of vectors in a basis for V. Notation is dim V. The dimension of the trivial vector space { 0 } is defined as 0. Dimension of a Vector Space
- 10. Corollary - If a vector space V has dimension n, then a largest linearly independent subset of vectors in V contains n vectors and is a basis for V Corollary - If a vector space V has dimension n, the smallest set of vectors that spans V contains n vectors and is a basis for V. Basis and Dimension
- 11. Corollary - If vector space V has dimension n, then any subset of m > n vectors must be linearly dependent. Corollary - If vector space V has dimension n, then any subset of m < n vectors cannot span V. Theorem - If S is a linearly independent set of vectors in a finite dimensional vector space V, then there is a basis T for V, which contains S. Basis and Dimension
- 12. Theorem - Let V be an n-dimensional vector space a) If S = { v1, v2, , vn } is a linearly independent set of vectors in V, then S is a basis for V. b) If S = { v1, v2, , vn } spans V, then S is a basis for V. Basis and Dimension
- 13. Consider the homogeneous system Ax = 0 where A reduces to Basis for Solution Sets : Homogeneous 1 2 2 2 3 0 4 6 4 6 6 4 1 3 4 2 4 0 2 4 4 3 4 2 A 1 0 2 0 3 4 0 1 2 0 1 0 0 0 0 1 2 2 0 0 0 0 0 0
- 14. Corresponding system of equations is 51 3 6 52 3 54 6 2 3 4 0 2 0 2 2 0 x x x x x x x x x x 1 2 3 4 5 6 2 3 4 2 3 4 2 2 1 0 1 0 0 2 2 0 2 2 0 1 0 0 0 1 x r s t x r s x r r s t x s t x s x t Let x6 = t, x5 = s, x3 = r Then x4 = –2s + 2t, x2 = –2r – s, x1 = 2r + 3s – 4t Basis for Solution Sets : Homogeneous
- 15. The null space of A is spanned by the independent set of 3 vectors (below) and thus has dimension 3. Dimension of null space is called the nullity of A. 2 3 4 2 1 0 1 0 0 , , 0 2 2 0 1 0 0 0 1 Basis for Solution Sets : Homogeneous
- 16. Consider the linear system The solution consists of all vectors of the form for arbitrary r and s Basis for Solution Sets : Nonhomogeneous 1 2 3 1 2 1 3 2 4 2 6 3 6 3 9 x x x 1 2 3 1 2 1 2 1 1 1 1 0 2 2 0 1 x r s x r r s x s
- 17. The set of linear combinations, , forms a plane passing through the origin of R3. This plan is a linear space. Basis for Solution Sets 2 1 1 0 0 1 r s
- 18. The solution vectors, , form a plane parallel to the plane above, displaced from the origin by the vector This plan is not a linear space. Basis for Solution Sets 1 2 1 1 1 0 2 0 1 r s 1 1 2
- 19. Section 4.4 p.259, p.260 1 to 10; 12 to 20; Problems