Chap 2.II.1 - Linear Dependence/Independence
MTH 214 - Linear Algebra
Dr. Lew
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The spanning set for a subspace is collection of vectors from which all
vectors of the subspace are created using linear combinations.
For example, each of the following are spanning sets for the vector space
P1 (R) = {a + bx : a, b ∈ R}:
{1, x} ,
{1, 2x, 5 − 6x}
Notice that one contains two vectors while the other one contains three.
This leads to a question: “What is the minimal number of vectors needed to
span a vector space?”.
Closer inspection of the spanning set {1, 2x, 5 − 6x}, there seems to be
some “redundancy” in the vectors listed.
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What we mean is that notice the third polynomial is actually a linear
combination of the first two: 5 − 6x = 5(1) + (−3)(2x).
As a consequence, any linear combination of all three of these vectors can be
rewritten as a linear combination of the two vectors. For example,
4(1) + 2(2x) − 2(5 − 6x) = 4(1) + 2(2x) − 2[5(1) + (−3)(2x)]
= 4(1) + 2(2x) − 10(1) + 6(2x)
= −6(1) + 8(2x)
If we remove the vector 5 − 6x from the spanning set, the remaining
two vectors would still span the same subspace. Also notice that the
remaining two vectors are not scalar multiples of each other.
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The “redundancy” we see in this spanning set is the observation that one
of the vectors in the spanning set is a linear combination of the other two
vectors.
Linear Dependence
Let S be a set of vectors in a vector space V .
We say that S is a linearly dependent set if some vector in S is a linear
combination of the other vectors in S.
A set of vectors that is not linearly dependent is called linearly independent.
Before we consider some consequences of this definition, let’s look at an
example.
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Example
Consider the vectors {v1 , v2 , v3 } in R3 , where
 
 
 
1
4
2
 
 
 
v1 = 2 , v2 = 5, v3 = 1
3
6
0
Check that v2 = 2v1 + v3 , so {v1 , v2 , v3 } is a linearly dependent set.
This means that any linear combination of the vectors v1 , v2 , v3 can be
rewritten as a linear combination of just v1 , v3 since any occurrence of
v2 in an expression can be replaced with 2v1 + v3 .
In terms of spanning sets, we see that Span{v1 , v2 , v3 } = Span{v1 , v3 }.
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Notice that we can rewrite v2 = 2v1 + v3 as
2v1 − v2 + v3 = 0
This leads to another view of linear dependence.
Lemma
A set S = {v1 , v2 , . . . , vk } in a vector space V is linearly dependent if
and only if there are scalars c1 , c2 , . . . , ck not all zero such that
c1 v1 + c2 v2 + · · · + ck vk = 0
Equivalently, we say S = {v1 , v2 , . . . , vk } is linearly independent if and
only if the only solution to the equation
x1 v1 + x2 v2 + · · · + xk vk = 0
is x1 = x2 = · · · = xk = 0.
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So if S is a spanning set for a subspace W of a vector space V , then
If S is a linearly dependent set, then one of the vectors in S is a linear
combination of the other vectors in S.
So we can remove this vector from S. The remaining set of vectors
S1 is still a spanning set for W .
Now look at S1 . Either it is a linearly independent set, or it isn’t. If
it isn’t, then there is some vector in S1 that is a linear combination
of the other vectors in S1 .
Remove this vector from S1 . The remaining set of vectors S2 is still
a spanning set for W .
By repeating this process, eventually we arrive at a set of vectors
that spans W , but is linearly independent.
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