MATH136 Lecture Notes - Lecture 20: Linear Independence, Spain, Linear Combination
14 May 2018
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Wednesday, June 14
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Lecture 20: Linear independence (Refers to 4.1)
Concepts:
1. Linearly independent set.
2. Recognize that subsets of linearly independent sets are linearly independent.
3. Characterize a linearly independent set as one being a set where no vector is a
linear combination of the others.
4. Unique representation property
20.1 Definition – Generalization of the definition of linear independence. Let v1, v2, ...., vk
be k vectors in a vector space V. The vectors v1, v2, ..., vk are said to be linearly
independent if and only if the only way that
α1v1 + α2v2 + ... + αkvk = 0
can hold true is if α1, α2, ....., αk are all zeroes.
If v1, v2, ..., vk are not linearly independent (i.e. there exists α1, α2 , ....., αk not all zero
such that α1v1 + α2v2 + ..... + αkvk = 0 ) then they are said to be linearly dependent.
20.1.1 Example – We know that the two functions ex and sin x are vectors in the set F
described above. Then S = Span{ex, sin x} = { αex + βsin x : α, β belong to ℝ } is a
subspace of F. Show that the set {ex, sin x} is linearly independent.
Solution: Let us first recall that two functions f and g are equal on their domain D if
and only if f(x) = g(x) for all x in D. Suppose
αex + βsin x = 0(x)
for all x (where 0(x) denotes the zero function; it maps every x in the domain to 0.).
- Suppose there exists α ≠ 0 such that αex + βsin x = 0(x).
- Then ex = (–β/α)sin x for all x. This includes the value x = π.
- But 0 ≠ eπ = (–β/α)sin π = 0, a contradiction.
- Then α must be 0.
- Suppose β ≠ 0 is such that, 0ex + βsin x = 0(x). Then (–0/β)ex = sin x for all x.
- Since sin(π/2) =1 ≠ 0, we have a contradiction.
- So β must also be 0.
- So {ex, sin x} is linearly independent.
20.1.2 Proposition − The vectors M = {v1, v2 , ..., vk} , with k ≥ 2, are linearly
independent in a vector space V if and only if no vector in M is a linear combination of
the other vectors in M.
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