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MATH 136
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Robert Sproule
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Lecture 20

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Mathematics

MATH 136

Robert Sproule

Winter

Description

Wednesday, February 26 − Lecture 20: Linear independence
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 isa
linear combination of the others.
20.1 Definition – Generalization of the definition of linear independence. Let v , v ...., v 1 2, k
be k vectors in a vector space V. The vectors v , v ..., v are said to be linearly
1 2, k
independent if and only if the only way that
α 1 1 α v 2 2.. + α v = 0k k
can hold true is if α ,1α ,2....., α kre all zeroes.
If v1, v2,.., vkare not linearly independent (i.e. there existsα , α , .1...2 α not alk zero
such that α v1 1α v +2 2... + α v = k k then they are said to be linearly dependent.
x
20.1.1 Example – We know that the two functions e and sinx are vectors in the set F
described above. Then S = Span{e , sinx} = { αe + βsinx : α, β belong to ℝ } is a
x
subspace of F. Show that the set {e , sinx} is linearly independent.
Solution: Let us first recall that two functions f and g are equal on their domain Dif
and only if f(x) = g(x) for all x in D. Suppxse
αe + βsinx = 0(x)
for all x (where 0(x) denotes the zero function; it maps every x in the domain to 0.).
x
- Suppose there exists α ≠ 0 such that αe + βsinx = 0(x).
x
- Then e = (–β/α)sinx for all x. This includes the value x = π.
- But 0 ≠ e = (–β/α)sinπ = 0, a contradiction.
- Then α must be 0.
x x
- Suppose β ≠ 0 is such that, 0e + βsinx = 0(x). Then (–0/β)e = sinx for all x.
- Since sin(π/2) =1 ≠ 0, we have a contradiction.
- So β must also be 0.
- So {e , sinx} is linearly independent.
20.1.2 Proposition − The vectors M = {v , v , ..1, v2} with k , 2, are linearly
independent if and only if no vector in M is a linear combination of the other vectors
in M.
Proof : This is similar to the previously given proof for the case of V = ℝ . n (⇐)
- Suppose none of these vectors is a linear combination of the others.
- Suppose M is not linearly independent.
- Then there exists α , α 1...2., α not kll zeroes such that α v + α v +1 1..+ α2 2= k k
0.
- Suppose α is not zero. Rearrange the order of {v , v , ...., v } so that p = 1, .i.e.,
p 1 2 k
α 1s not zero.
- Then v = 1 (α /α )2v +1− (2 /α ) v +3, 1.., 3 − (α /α )v . k 1 k
- Hence v is a linear combination of the others. Contradiction.
1
(⇒)
- Suppose {v , v1, .2.., v } ks linearly independent.
- Suppose one vector of {v , v , .1..,2v } is k linear combination of the others. Say it
is v .
1
- Then there exists α ....2, α suck that such that v = α v + 1.... 2 2 v . k k
- Then v − 1 v − 2 v2+ , ..3, 3 α v . = 0. k k
- Since the coefficient of v is n1t zero this contradicts our hypothesis.
- Then no vector of {v , v ,1...2, v } iska linear combination of the others.
If a vector in U is a linear combination of the others we will refer to it as being
“redundant”. It doesn't contribute anything to its span.
20.2 Definition – Let {v 1, v2,..., vk} be a spanning family for a vector space V. That is,
Span{v , v ..., v } = V. Then for every vector v in V there exists scalars α , α , ....., α
1 2, k 1 2 k
such that
v = α v1 1α v +2 2... + α v . k k
If for each vectorv this set of scalars α , α , ..., α satisfying v = α v + α v + ..... + α v
1 2 k 1 1 2 2 k k
is unique then we say that Span{v 1, v2,.., v k satisfies the unique representation
property with respect to its spanning family.
20.2.1 Example – The spanning family Span{(1, 0), (0,1)} = Span{ e , e } = ℝ 1 2 2
satisfies the unique representation property since for an arbitrary vector v= (a, b) in
ℝ , a and b are the unique scalars associated to v so that v = ae + be . 1 2
20.2.2 Theorem − Let S = {v , v , .1..,2v } be a k

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