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Linear dependence of vectors

  1. Aug 28, 2014 #1
    Hi all,

    I was asked by someone today to explain the notion of linear independence of a set of vectors and I would just like to check that I explained it correctly.

    A set of vectors [itex] S[/itex] is said to be linearly dependent if there exists distinct vectors [itex] \mathbf{v}_{1}, \ldots , \mathbf{v}_{m} [/itex] in [itex] S[/itex] and scalars [itex] c_{1},\ldots c_{m}[/itex], not all of which are zero, such that [tex] c_{1}\mathbf{v}_{1} + \cdots + c_{m}\mathbf{v}_{m} = \sum_{i=1}^{m} c_{i}\mathbf{v}_{i} = \mathbf{0} [/tex]

    What this means is that at least one vector in [itex]S[/itex] can be completely specified in terms of the other vectors in the set and hence it is dependent on the particular form of those vectors. However, if the only case for which [itex]\sum_{i=1}^{m} c_{i}\mathbf{v}_{i} = \mathbf{0}[/itex] is the trivial case, in which [itex] c_{i} = 0 \; \forall \; i=1, \ldots , m[/itex], then the set is said to be linearly independent, as none of the vectors contained within it can be specified in terms of the other vectors in [itex]S[/itex].

    Is this a valid description of the concept?
     
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  3. Aug 28, 2014 #2

    mathman

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    Minor quibble. Usually [itex]S=(\mathbf{v}_{1}, \ldots , \mathbf{v}_{m}) [/itex]
     
  4. Aug 28, 2014 #3

    micromass

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    Yes, I think that is a very good way of describing it.
     
  5. Aug 28, 2014 #4

    Fredrik

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    Edit: Never mind, I read the OP wrong.
     
    Last edited: Aug 28, 2014
  6. Aug 28, 2014 #5

    WWGD

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    Do you mean the basis is ordered?
     
  7. Aug 28, 2014 #6

    micromass

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    I don't see how this is correct. For example, ##S## can be infinite and then your correction doesn't apply. The definition in the OP is entirely correct.
     
  8. Aug 28, 2014 #7
    Thanks for your help on the matter guys, much appreciated.
     
  9. Aug 28, 2014 #8
    Also, in addition is this reasoning correct for proving that the representation of a given vector [itex]\mathbf{v}[/itex] in a vector space [itex]V[/itex], with respect to a given basis [itex]\mathcal{B}=\lbrace \mathbf{e}_{i} \rbrace_{i=1, \ldots , n} [/itex], is unique? :

    Let [itex]V[/itex] be an [itex]n[/itex]-dimensional vector space and [itex]\mathcal{B}=\lbrace \mathbf{e}_{i} \rbrace_{i=1, \ldots , n} [/itex] be a given basis for [itex] V[/itex]. Suppose that a given vector [itex]\mathbf{v} \in V [/itex] can be represented in terms of the basis [itex]\mathcal{B}[/itex] as two linear combinations [itex] \sum_{i=1}^{n}a_{i} \mathbf{e}_{i} [/itex] and [itex] \sum_{i=1}^{n}b_{i} \mathbf{e}_{i} [/itex]. Then, [tex] \sum_{i=1}^{n}a_{i} \mathbf{e}_{i} = \sum_{i=1}^{n}b_{i} \mathbf{e}_{i} [/tex] such that [tex] \sum_{i=1}^{n} \left( a_{i}-b_{i} \right) \mathbf{e}_{i} = \mathbf{0} [/tex] As the vectors [itex] \lbrace \mathbf{e}_{i} \rbrace_{i=1, \ldots , n} [/itex] form a basis they are, by definition, linearly independent. Hence, this implies that [itex] a_{i} = b_{i} \; \forall \; i= 1, \ldots n [/itex]. The coefficients [itex] a_{i}[/itex] and [itex] b_{i}[/itex] must satisfy the condition that their linear combination (along with the basis vectors) describe the vector [itex] \mathbf{v} [/itex], but are otherwise arbitrary, and hence we must conclude that in fact there is only one, unique, set of scalars [itex] \lbrace a_{i} \rbrace [/itex] that satisfy [itex] \mathbf{v} = \sum_{i=1}^{n}a_{i} \mathbf{e}_{i} [/itex].
     
  10. Aug 28, 2014 #9

    Fredrik

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    Yes, that proof is fine.

    If you had been using a definition of "basis" that doesn't make it clear that every basis is a linearly independent set, you would also have had to prove linear independence.
     
  11. Aug 28, 2014 #10
    Thanks. Is the argument I gave at the end about why [itex] a_{i} =b_{i} [/itex] for arbitrary scalars [itex] a_{i}, b_{i} [/itex], satisfying the required properties, requires that there is actually only one set of scalars?
     
  12. Aug 28, 2014 #11

    Fredrik

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    Yes, this is the standard way to prove uniqueness. If you know that x has property P, and you want to prove that nothing else does, you show that for all y with property P, we have y=x.
     
  13. Aug 28, 2014 #12
    Ok, cool. Thanks for your help.
     
  14. Aug 29, 2014 #13

    mathman

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    The statement talks about one particular set of vectors. In the case S is infinite, the statement would have to say for every finite set of basis vectors.
    -------------------------------------------------------------------

    No. I just meant S was the given set. In general, when talking about vector spaces, ordering is not relevant.
     
  15. Aug 29, 2014 #14

    micromass

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    The OP never even mentioned basis vectors :confused:
     
  16. Aug 30, 2014 #15

    mathman

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    You are right. However his question was about a particular (finite) set of vectors, which was the question I was addressing.
     
  17. Aug 30, 2014 #16

    WWGD

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    -------------------------------------------------------------------

    No. I just meant S was the given set. In general, when talking about vector spaces, ordering is not relevant.[/QUOTE]

    You're right for f.d case, or for Hamel bases, where sums are finite, which is the case here. I know this is absurdly far-off for this post, but for the sake of a broader context , order does matter for Schauder bases, where convergence is conditional. And order matters when defining an iso. between linear maps and their representation as matrices.
     
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