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Can you rearrange vectors in a set? And another misc questn.

  1. Jan 28, 2016 #1
    Suppose you have a set of vectors v1 v2 v3, etc.

    However large they are, suppose they span some area, which I think is typically represented by

    Span {v1, v2, v3}
    But I mean, if you're given these vectors, is there anything wrong with rearranging them? Because there's a theorem- that
    "an indexed set S= {v1, v2... vp} of more than one vectors is linearly dependent if at least one vector is in a linear combination of the others."
    So if S is linearly dependent, any vector in the set is a combination of the preceeding vectors?
    Or did I read that wrong, and it just means a certain vector, possibly more than one is a lin comb of some other vectors?

    However the theorem i'm reading seems to really detail that there's something special about "preceeding vectors". So if you have any set, is interchanging vectors allowed?
    I feel like that there is nothing wrong with this. Is there some time when this is allowed and it isnt, maybe?
    (I've just started linear algebra for a few weeks so I don't know any complex scenarios)

    But it seems that this theorem suggests that there's something important to the permutation of these vectors.
  2. jcsd
  3. Jan 28, 2016 #2


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    Reordering the vectors in a spanning set has no effect. There's nothing wrong with it.

    When we talk about a vector space basis, we may wish to imply an ordering, because without an ordering, we cannot speak unambiguously of the representation of a vector in that basis, which we often wish to do. If we take that definition of 'basis' then, for every set of linearly independent, spanning vectors in an n-dimensional vector space, there are n! different normalized bases, corresponding to the number of ways the vectors could be reordered.

    My guess is that the reference to 'preceding' is just about the method by which one tests linear independence. One way to do that is to label the vectors as v1, v2, ... , vn. Then test that v2 is independent of v1, Next test that v3 is independent of v1 and v2, and so on. But that ordering is just a convenience used in performing the test, not an intrinsic requirement of the set.
  4. Jan 28, 2016 #3
  5. Jan 28, 2016 #4
    But for instance, does this mean if you try to solve a matrix of [v1 v2 v3] and a matrix with just rearranged vectors like [v3 v1 v2]......... it's the same??
  6. Jan 28, 2016 #5


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    By 'solve a matrix' do you mean calculate its (multiplicative) inverse? If so then, no, the answer is not the same.
  7. Jan 28, 2016 #6
    Ummmm I'm not sure.

    Does it make a difference how you solve it?

    For instance I've only learned about Ax=b, using the matrix A as a function. And also solving span{v1 v2 v3}=0, to test for interdependence.

    I don't know about what the inverse is.

    But maybe I meant if you switch the positions of vectors in a set, isn't that equivalent to swapping collumns in a matrix? In that case, then a matrix of [v1 v2 v3] is equivalent in any respect to [v3 v1 v2]?
  8. Jan 28, 2016 #7


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    It's equivalent in the sense that
    $$[v3\ v1\ v2] =
    [v1\ v2\ v3]
    \left( \begin{array}{ccc}
    0 & 1 & 0 \\
    0 & 0 & 1 \\
    1 & 0 & 0 \end{array} \right)$$
    [Or something like that. I often get my rows and cols muddled up in matrix mults]

    Equation Ax=b will have a completely different solution from A*x=b where A* is A with shuffled columns.
    However it will have the same solution as A*x=b*, where b* is b with the same shuffle applied to its components as was applied to the columns of A to get A*.
  9. Jan 29, 2016 #8
    Oh alright. Thanks.
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