Results 1 to 11 of 11

Thread: permuting vectors

  1. #1
    Newbie
    Joined
    May 2006
    Posts
    2

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jun 2005
    Posts
    295
    Awards
    1
    The answer is "it depends". Consider, for example, the permutations of (1,1,1,...1) and the permutations of (1,0,0,...,0).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    Rgep's examples suggest to me this conjecture: The dimension of the subspace is 0 if $\displaystyle v = (0,0, \ldots ,0)$, 1 if $\displaystyle v = (a,a, \ldots ,a)$ for some $\displaystyle a \ne 0$, and $\displaystyle n$ otherwise. Suggestions for a proof or counter-examples?
    Last edited by JakeD; May 2nd 2006 at 01:25 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    Well, that seems to be the case. The proof is really straightforward, let $\displaystyle v$ have a nonzero entry a at the i-th place: $\displaystyle v=(v_1,...,a,...,v_n)$. We can take $\displaystyle a=1$, as otherwise we can consider $\displaystyle (1/a)v$.

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

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

    Then

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

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

    This proves $\displaystyle V$ contains the j-th axis of $\displaystyle R^n$. So $\displaystyle V$ contains every axis, and therefore $\displaystyle V=R^n$.
    Last edited by Rebesques; May 3rd 2006 at 03:43 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Generalizing what rgep said, it seems to me I was (unable to prove it).

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

    Otherwise its dimension is $\displaystyle n$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    -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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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 $\displaystyle
    v = (0,0, \ldots ,0)
    $
    You sure about that? It looks like the dimension is one to me.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    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 $\displaystyle \mathbb{R}^n$ of dimension 0.

    RonL
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    My house.
    Posts
    658
    Thanks
    42
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by CaptainBlack
    No the space consists of a single point - a subspace of $\displaystyle \mathbb{R}^n$ 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.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Senior Member
    Joined
    Apr 2006
    Posts
    399
    Awards
    1
    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 $\displaystyle v = (x_1, x_2,\ldots, x_n).$ The dimension $\displaystyle d$ of the subspace $\displaystyle V$ spanned by $\displaystyle v$ and its permutations is

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

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

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

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

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

    Putting these vectors into a matrix $\displaystyle \begin{bmatrix}
    1 & 0 & 0 & 0 & -1 \\
    0 & 1 & 0 & 0 & -1 \\
    0 & 0 & 1 & 0 & -1 \\
    0 & 0 & 0 & 1 & -1 \\
    x_1 & x_2 & x_3 & x_4 & x_5 \\
    \end{bmatrix}
    $
    it is clear that the first $\displaystyle n -1 = 4$ 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 $\displaystyle n - 1$ vectors and the number of independent vectors in $\displaystyle V$ will be $\displaystyle n,$ 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 $\displaystyle n - 1$ vectors. Since the last vector could represent any permutation of $\displaystyle v$, this means there cannot be more than $\displaystyle n - 1$ independent vectors in $\displaystyle V.$ This proves C and completes the proof.
    Last edited by JakeD; May 6th 2006 at 08:06 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: Nov 15th 2011, 05:10 PM
  2. Replies: 3
    Last Post: Jun 30th 2011, 08:05 PM
  3. Replies: 2
    Last Post: Jun 18th 2011, 10:31 AM
  4. [SOLVED] Vectors: Finding coefficients to scalars with given vectors.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Jan 23rd 2011, 12:47 AM
  5. Permuting roots....
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Apr 4th 2009, 12:18 PM

Search Tags


/mathhelpforum @mathhelpforum