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Spanning sets, and linear independence of them

  1. Jan 24, 2007 #1
    I've become sort of confused on the topic of the linear span versus spanning sets. I know that the span of a subset is the set containing all linear combinations of vectors in V. Is a spanning set then the same thing, or is it something else?

    Also, in terms of bases... A basis is a linearly independent spanning set, but I thought a span was a set containing linear combinations... BUT linear combinations generally indicate linear dependence! If that's the case, how is the spanning set linearly independent? I know I'm missing something here, just not sure what! Anyone have a good description that might help? :shy:
  2. jcsd
  3. Jan 24, 2007 #2
    I haven't heard of a spanning set, so I'll use linear span to denote the set of all finite linear combinations of (a finite number of) vectors belonging to some vector space V. It is evident that this set is a subspace of V.

    A basis of V is a set of linearly-indepedent vectors (that belong to V), which span the whole of V. If you consider the linear span of the basis vectors, it would contain all the vectors in V.

    Basis : {e1, e2, e3,...,en} - these vectors are linearly-independent and belong to V. Every vector v, belonging to V, can be expressed as a linear combination of these vectors.
  4. Jan 24, 2007 #3


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    "span" and "spanning set" are, in a sense, opposites. Given a collection of vectors {v1,v2, ... , vn}, the set of all possible linear combinations of those, {a1v1+ a2v2+ ... + anvn} is its "span".

    Conversely, if U is a subspace of V, a collection of vectors that has U as its span is a "spanning set" for U.

    ?? Where did you get that idea? A set of vectors is independent if and only if the only way you can have a linear combination of them equal to the 0 vector is if all the coefficients in the combination are 0. It is easy to prove from that that each vector in their span can be written as a linear combination of them in only one way. In, for example, R3, the two vectors <1, 0, 0> and <0, 1, 0> are independent. There span is the set of all vectors of the form a<1, 0, 0>+ b<0, 1, 0>= <a, b, 0> but they themselves are independent.

    In a sense the concepts of "spanning" and "independent" are opposites. A set containing a single vector is obviously "independent". As you add more vectors it becomes more likely that it becomes dependent. On the other hand, a set containing all vectors clearly spans the entire space. As you remove vectors it becomes more likely that you will miss one. The crucial fact for (finite-dimensional) vectors is this: In order for a set of vectors in an n-dimensional space to be independent there cannot be more than n vectors in the set. In order for a set of vectors to span the space there cannot be less than n vectors in the set. In order to be both "spanning" and "independent", there must be exactly n vectors in the set: every basis of an n-dimensional space contains n vectors.
    Last edited by a moderator: Jan 24, 2007
  5. Jan 24, 2007 #4
    that makes a lot more sense. I guess I was forgetting that linear combinations can be either dependent or independent, but both are a possibility... so if the combination contains the zero vector or all the vectors are zero, then it must be dependent, and otherwise independent.

    and I guess then that a linear span can be of a combination which is dependent or independent. :rofl:
  6. Jan 24, 2007 #5


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    I'm not sure that the term 'dependent/independent linear combination' makes sense. A linear combination can consist of dependent or independent vectors, if that's what you meant.

    If the combination contains the zero vector, then it consists of dependent vectors, since that vector is dependent with any other vector, i.e. a set containing the zero vector is dependent.
  7. Jan 24, 2007 #6


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    Agree with radou- it isn't the linear combination that is "independent" or "dependent", it is the set of vectors- and they can be involved in many linear combinations.
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