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4.3 Linearly Independent Sets McDonald Fall 2018, MATH 2210Q, 4.3 - - PDF document
4.3 Linearly Independent Sets McDonald Fall 2018, MATH 2210Q, 4.3 - - PDF document
4.3 Linearly Independent Sets McDonald Fall 2018, MATH 2210Q, 4.3 Slides 4.3 Homework : Read section and do the reading quiz. Start with practice problems. Hand in : 3, 4, 14, 21, 29, 30 Recommended: 8, 10, 15, 23, 24, 31 Definition 4.3.1.
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Example 4.3.10. Verify B = {1, t, t2, · · · , tn} is a basis for Pn. This is the standard basis for Pn.
4.3.1 The spanning set theorem
Example 4.3.11. Let v1 = 2 −1 , v2 = 2 2 , v3 = 6 16 −5 , and H = Span{v1, v2, v3}. Verify that v3 = 5v1 + 3v2, and Span{v1, v2, v3} = Span{v1, v2}. What is a basis for H? Definition 4.3.12. Let S = {v1, · · · , vp} be a set in V , and let H = Span{v1, · · · , vp}. (a) If one of the vectors in S, say vk, is a linear combination of the remaining vectors in S, then the set formed by removing vk from S still spans H. (b) If H = {0}, some subset of S is a basis for H. 3
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4.3.2 Bases for Col A and Nul A
Example 4.3.13. Find a basis for Col U, where U =
- u1
· · · u5
- =
1 4 2 1 −1 1 . Example 4.3.14. Below, A is row equivalent to U from the last example. Find a basis for Col A. A =
- a1
· · · a5
- =
1 4 2 −1 3 12 1 5 5 2 8 1 3 2 5 20 2 8 8 . Theorem 4.3.15. The pivot columns of a matrix A form basis for Col A. Watchout! 4.3.16. We need to reduce A to echelon form U to find pivot columns. However, the pivot columns of U do not form a basis for Col A. You have to use the pivot columns of A. 4
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Example 4.3.17. Find a basis for Nul A, where A is the same as the previous example: A = 1 4 2 −1 3 12 1 5 5 2 8 1 3 2 5 20 2 8 8 .
4.3.3 Two views of a basis
Example 4.3.18. Which of the following is a basis for R3? 1 , 2 3 1 , 2 3 , 4 5 6 1 , 2 3 , 4 5 6 , 7 8 9 Remark 4.3.19. In one sense, a basis for V is a spanning set of V that is as small as possible. In another sense, a basis for V is a linearly independent set that is as large as possible. 5
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