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Vector

Vector. Linear Combination and Linear Independence. Linear Combination. An expression of the form a 1 v 1 + a 2 v 2 + … + a n v n is called a linear combination of v 1 , v 2 , … v n. Question 1:. If a = 3 i + 2 j, b = -4 i + j and p = i + 8 j

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Vector

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  1. Vector Linear Combination and Linear Independence

  2. Linear Combination • An expression of the form a1v1 + a2v2 + … + anvn • is called a linear combination of v1, v2, …vn.

  3. Question 1: • If a = 3i + 2j, b = -4i+j and p = i + 8j • Express p in terms of a and b

  4. Solution: • Let p = ma + nb, • then i + 8j = m(3i + 2j) + n(-4i + j). • i + 8j = (3m – 4n)i + (2m + n)j • (1) + 4(2): 11m = 33 • m = 3, n = 2 • Therefore, p = 3a + 2b.

  5. p 4b 3a Meaning of linear combination b a

  6. Question 2: • If a = i–j + 3k, b = 2i- k p = 3i + j – 5k • Express p in terms of a and b

  7. Solution: • If p = ma + nb, then • 3i + j– 5k = m(i–j + 3k) + n(2i - k). What if p = 3i + j + 2k? • From (2): m = -1 • From (1): n = 2 (also satisfying (3)) • Therefore, p = -a + 2b.

  8. Observation from Q.2 • If we only use two vectors a and b, we will NOT be able to ‘generate’ all vectors p in R3. • How about 3 vectors? • Of course, i, j and k can ‘generate’ all vectors p in R3. • How about a = i + j, b = j + k and c = i–k?

  9. a = i + j b = j + k c = i–k p = i–j + k If p = xa + yb + zc, then: (1) + (3): x + y = 2, which contradicts (2). Try expressing p in terms of a, b and c !

  10. a = i + j b = j + k c = i–k These 3 vectors are NOT able to ‘generate’ all vectors p in R3. Actually, a = b + c: a c b a also lies on this plane because it is a linear combination of b and c. Conclusion

  11. Linear Dependence • A set of vectors v1, v2, …, vn are called linearly dependent if one of them is a linear combination of others. • Otherwise, they are called linearly independent. • Can you give an example of 3 linearly independent vectors? • How about i + j, 0, k– 2i ?

  12. Main Theorem • v1, v2, …, vn are linearly dependent iff  scalars a1, a2, …, an, not all zero, s. t. a1v1 + a2v2 + … + anvn= 0.

  13. Proof of the “if” part () • If  scalars a1, a2, …, an s. t. ai  0 and a1v1 + a2v2 + … + anvn= 0. • vi is a linear combination of others. • Hence, the vectors are linearly dependent.

  14. Proof of the “only if” part () • If one of the vectors, say, vi is a linear combination of others, then • vi = a1v1 + … + ai - 1vi - 1 + ai + 1vi + 1 + … + anvn for some a1, …, ai – 1, ai+ 1, …, an. • a1v1 + … + ai - 1vi - 1 + (–1)vi + ai + 1vi + 1 + … + anvn = 0 • a1v1 + a2v2 + … + anvn= 0 where ai = -1. ai

  15. Main Theorem (2) • v1, v2, …, vn are linearly dependent iff  scalars a1, a2, …, an, not all zero, s. t. a1v1 + a2v2 + … + anvn= 0. • On the other hand, v1, v2, …, vn are linearly independent if and only if whenever a1v1 + a2v2 + … + anvn= 0, we have a1 = a2 = … = an = 0.

  16. Question: • Determine whether a = (1, 2, 3) b= (-2, 5, 2) c = (3, 1, -3) are linearly independent. i.e. If xa + yb + zc =0, where x, y, z R determine whether x, y and z are all zeros.

  17. Solution: • Consider xa + yb + zc = 0. Then, How do we call this kind of system? When will it has non-trivial solution?

  18. the homogeneous system only has the trivial solution. x = y = z = 0 a, b and c are linearly independent. Answer:

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