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Math Help - permuting vectors

  1. #1
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    permuting vectors

    Let v=(x[1],...,x[n]) be a n-dimensional real vector. Find the dimension of the subspace of R^n spanned by v and all vectors obtained by permuting the components of v.
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    The answer is "it depends". Consider, for example, the permutations of (1,1,1,...1) and the permutations of (1,0,0,...,0).
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    Rgep's examples suggest to me this conjecture: The dimension of the subspace is 0 if v = (0,0, \ldots ,0), 1 if v = (a,a, \ldots ,a) for some a \ne 0, and n otherwise. Suggestions for a proof or counter-examples?
    Last edited by JakeD; May 2nd 2006 at 01:25 AM.
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    Well, that seems to be the case. The proof is really straightforward, let v have a nonzero entry a at the i-th place: v=(v_1,...,a,...,v_n). We can take a=1, as otherwise we can consider (1/a)v.

    Let V be the subspace created by v and its permutations, and consider the j-axis {(0,..,x_j,...,0)} of R^n.

    We consider a permutation s of \{1,2,...,n\} with 1 occupying the j-th place, and denote the vector obtained by v^*=(v_{s(1)},...,1,...,v_{s(n)}).

    Then

    v^*=(v_{s(1)},...,v_{s(j-1)},0,v_{s(j)},...,v_{s(n)})+{\bf e}_j

    where {\bf e}_j is the jth euclidean basis vector. And
    (0,..,x_j,...,0)=x_j{\bf e}_j+0(v_{s(1)},...,v_{s(j-1)},0,v_{s(j)},...,v_{s(n)})

    This proves V contains the j-th axis of R^n. So V contains every axis, and therefore V=R^n.
    Last edited by Rebesques; May 3rd 2006 at 03:43 AM.
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    Generalizing what rgep said, it seems to me I was (unable to prove it).

    That there are only two cases, given,
    (x_1,x_2,...,x_n)
    And if x_1=x_2=...=x_n\not = 0
    Then, the dimension of this subspace is 1.

    Otherwise its dimension is n
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    Super Member Rebesques's Avatar
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    -Doctor, i feel people don't pay attention to me.

    -Who's next please?
    Last edited by Rebesques; May 2nd 2006 at 05:04 PM.
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    Sorry, 'bout that did not wish to steal anyone's answers I just found the problem interesting.

    Quote Originally Posted by JakeD
    The dimension of the subspace is 0 if <br />
v = (0,0, \ldots ,0)<br />
    You sure about that? It looks like the dimension is one to me.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker
    Sorry, 'bout that did not wish to steal anyone's answers I just found the problem interesting.


    You sure about that? It looks like the dimension is one to me.
    No the space consists of a single point - a subspace of \mathbb{R}^n of dimension 0.

    RonL
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    Super Member Rebesques's Avatar
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    Sorry, 'bout that did not wish to steal anyone's answers
    Oh no no, I made no such claim

    I just felt invisible for a moment there
    Last edited by Rebesques; May 3rd 2006 at 07:48 AM.
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    Quote Originally Posted by CaptainBlack
    No the space consists of a single point - a subspace of \mathbb{R}^n of dimension 0.

    RonL
    You are right. Understand my mistake, I was considering that the set of all unit vectors for a basis and are linearly independent, my fault was that they were not elements from the zero vector.
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    In working up a proof of my conjecture, I found a counter example and had to revise the conjecture. (Rebesques, take note. :( ) Here is the revised conjecture and a proof.

    Let v = (x_1, x_2,\ldots, x_n). The dimension d of the subspace V spanned by v and its permutations is

    \begin{array}{lll}<br />
\mathsf{A.} &0 &\mathsf{if}\ v = (0,0,\ldots,0), \\<br />
\mathsf{B.} &1 &\mathsf{if}\ v = (a,a,\ldots,a) \ \mathsf{for\ some}\ a \ne 0, \\<br />
\mathsf{C.} &n-1 &\mathsf{if}\ v \ne (a,a,\ldots,a) \ \mathsf{for\ any}\ a<br />
\ \mathsf{and}\ \sum_{i=1}^{n} x_i = 0, \\<br />
\mathsf{D.} &n &\mathsf{if}\ v \ne (a,a,\ldots,a) \ \mathsf{for\ any}\ a<br />
\ \mathsf{and}\ \sum_{i=1}^{n} x_i \ne 0. \\<br />
\end{array}<br />

    As examples in \mathbb{R}^2, for v = (1,1)\ \mathsf{or}\ (1,-1),\ d = 1, while for v = (1,0),\ d = 2.

    Claims C and D are proven by showing that the given number is the maximum number of independent linear combinations of v and its permutations. So assume v = (x_1, x_2,\ldots, x_n) \ne (a,a,\ldots,a) for any a. Under this assumption we may further assume wlog that
    x_1 \ne x_2 since we can permute v as we like. Let \alpha = 1/(x_1 - x_2).

    We will use the following linear combinations of v and its permutations.

    \begin{array}{lll}<br />
v_1 &=& \alpha (x_1,x_3,x_4,x_5,x_2) - \alpha (x_2,x_3,x_4,x_5,x_1) \\<br />
&=& (1,0,0,0,-1), \\<br />
v_2 &=& \alpha (x_3,x_1,x_4,x_5,x_2) - \alpha (x_3,x_2,x_4,x_5,x_1) \\<br />
&=& (0,1,0,0,-1), \\<br />
v_3 &=& \alpha (x_3,x_4,x_1,x_5,x_2) - \alpha (x_3,x_4,x_2,x_5,x_1) \\<br />
&=& (0,0,1,0,-1), \\<br />
v_4 &=& \alpha (x_3,x_4,x_5,x_1,x_2) - \alpha (x_3,x_4,x_5,x_2,x_1) \\<br />
&=& (0,0,0,1,-1), \\<br />
v_5 &=& v \\ &=& (x_1,x_2,x_3,x_4,x_5). \\<br />
\end{array}

    Putting these vectors into a matrix \begin{bmatrix}<br />
1   &  0  &  0  &   0 & -1 \\<br />
0   &  1  &  0  &   0 & -1 \\<br />
0   &  0  &  1  &   0 & -1 \\<br />
0   &  0  &  0  &   1 & -1 \\<br />
x_1 & x_2 & x_3 & x_4 & x_5 \\<br />
\end{bmatrix}<br />
    it is clear that the first n -1 = 4 vectors are linearly independent and their elements sum to zero. If the elements of the last vector do not sum to zero, it will be independent of the first n - 1 vectors and the number of independent vectors in V will be n, the maximum possible. This proves claim D.

    Proving C, if the elements of the last vector sum to zero, it will be linearly dependent on the first n - 1 vectors. Since the last vector could represent any permutation of v, this means there cannot be more than n - 1 independent vectors in V. This proves C and completes the proof.
    Last edited by JakeD; May 6th 2006 at 08:06 PM.
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