MATH136 Lecture Notes - Lecture 21: Identity Matrix, Linear Algebra, Nonfinite Verb
14 May 2018
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Friday, June 16
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Lecture 21 : Basis and dimension of a vector space (Refers to 4.2)
Concepts:
1. Define a basis.
2. Recognize that any two bases have the same number of elements.
3. Verify that a set is linearly independent. Verify that a set is a basis.
4. Define dimension of a vector space, infinite basis
21.1 Definition − If {v1, v2, ...., vk} is both
1. A spanning family of V and
2. Linearly independent
then we say that {v1, v2, ....,vk} is a basis of V .
21.2 Example − Show that any n linearly independent vectors in the vector space ℝn
forms a basis of ℝn.
Solution:
Given: {v1, v2, ...., vn} linearly independent .
Suffices to show: ℝn = span{v1, v2, ...., vn}.
- Let A = [v1 v2 ....,vn], a square n by n matrix
- Let v in ℝn. Suffices to show: Ax = v is consistent.
- Since {v1, v2, ....,vn} is linearly independent Ax = 0 only has the trivial solution.
- This implies that ARREF is the identity matrix.
- Then the system [A | v] row reduces to [ I | w] and so has a unique solution.
- Thus Col(A) = ℝn. And so {v1, v2, ....,vn} spans ℝn.
- So {v1, v2, ....,vn} is a basis of ℝn
21.2.1 Remark – The trivial subspace {0} of a vector space V is a special vector space
which does not have a basis.
21.3 Theorem
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Suppose B = {w1, w2, …, wn} is a basis of V and {v1, v2, … vk} is a
subset of V which contains k distinct non-zero vectors. If k > n then {v1, v2, … vk} can
not be linearly independent in V.
Proof : Since B = {w1, w2, …, wn} is a basis of V then for each i = 1 to k,
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Friday, june 16 lecture 21 : basis and dimension of a vector space (refers to 4. 2) Concepts: define a basis, recognize that any two bases have the same number of elements, verify that a set is linearly independent. Verify that a set is a basis: define dimension of a vector space, infinite basis. 21. 1 definition if {v1, v2, , vk} is both: a spanning family of v and, linearly independent then we say that {v1, v2, ,vk} is a basis of v . Given: {v1, v2, , vn} linearly independent . This implies that arref is the identity matrix. Let a = [v1 v2 ,vn], a square n by n matrix. 21. 2 example show that any n linearly independent vectors in the vector space n forms a basis of n. Suffices to show: n = span{v1, v2, , vn}. Suffices to show: ax = v is consistent. So {v1, v2, ,vn} is a basis of n.
