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MATH 136
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Robert Andre
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Lecture

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Mathematics

MATH 136

Robert Andre

Spring

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Wednesday, January 8 − Lecture 2 : Linear independence of vectors in ℝ . n
Concepts:
1. Linearlyindependent subset in ℝ . n
2. Characterization of a linearly independent set as one being a set where no vector
is a linear combination of the others.
3. A plane in ℝ . A hyperplane in ℝ . n
2.1 Definitions.
- Let v 1, v2,.., v ke k vectors in ℝ and suppose the vector 0 represents the zero vector
0 = (0, 0, …, 0). The vectors v , v 1...2,v areksaid to be linearly independentif the
only way that
can hold true is if α ,1α ,2..., α ake all zeroes. The solution where all the α‘s are i
zeros is called the trivial solution of this vector equation.
- If v , v ..., v are not linearly independent then they are said to be linearly
1 2, k
dependent.
Note that a linearly independent set cannot contain the zero-vector, 0. This fact follows
from the definition. In class, we will often abbreviate the words “linearly independent”
with the letters “L.I.”.
3
2.2 Example − Verify whether the set {v 1, v2, v3} in ℝ where v = (1, 1, 1), v = (02 1, 7)
and v 3 (0, 0, 3) is linearly independent.
Solution:
The given set is linearly independent.
m
2.3 Theorem − A subset of a finite linearly independent subset of ℝ is linearly
independent. Proof:
m
- Suppose M = {v , v , 1..,2v } is r linearly independentsubset of ℝ .
- Let S = {v , k1, .k2., v } km a non-empty subset of M. We claim that S is linearly
independent.
- Suppose not. Then there exists α , α , k1 k2 …, α km ,not all equal to zero, such that
α k1+k1 v + …k2 k2v = 0. km km
- Suppose, without loss of generality, α ≠ 0. Then k1
0v 1 … + (α v + αk1 k1… + αk2 k2+ ... + 0v =km.km r
where α ≠ k1
- This contradicts the fact that M = {v , v , ..1., 2 } is a rinearly independent subset of
m
ℝ .
- The source of the contradiction is our supposition that S is not linearly independent.
- So S = {v k1, vk2 ...., vkm is linearly independent, as required.
2.4 Theorem. An important characterization of linear independence − The vectors
M = {v , v , ...., v } of ℝ n with k ≥ 2, are linearly independent ifand only if no vector
1 2 k ,
in M is a linear combination of the others.
Proof of (⇐):
Given: None of these vectors in M is a linear combination of the others.
n
Required to show: That M is linearly independent subset of ℝ .
Let’s suppose that M is not linearly independent.
- Then there exists α , α 1 ...2 α not akl zeroes such that
- Suppose one of the coefficients, sayα is not zerop,Rearrange the order of {v , v , ...., 1 2
vk} so that p = 1, .i.e., α is 1ot zero.
- Then
- Hence v is1a linear combination of the others. Contradiction.
- So M is linearly independent.
Proof of (⇒) :
Given: M = {v 1, v2, ...., k } is linearly independent.
Required to show: That no vector in M is a linear combination of the others. - Suppose one vector of {v , v , 1...2 v } is k linear combination of the others. Say it is
v1. We claim that this will lead to a contradiction of our hypothesis.
- Then there exists α ...2., α sukh that such that
- Then
- This contradicts the fact that M is linearly independent.
- Then no vector of {v , v ,1....2 v } iska linear combination of the others.
2.5 Remark − The above theorem shows that we may have defined “linearly
independent” as follows:
n
"The set U = {v , v1...2,v } iska linearly independent of ℝ if and only if no vector
in U is a linear combination of the other vectors in U."
If a vector v jin U is a linear combination of the others we refer to v as

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