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Let $K$ be a field. Consider the vector space $K^n$ over the field $K$. Suppose $(a_1,a_2, ... ,a_n) \in K^n$. What is the dimension of the subspace generated by all the permutations of $(a_1,a_2,...,a_n)$?

I think there are 4 different cases

  1. $a_1=a_2=...=a_n=0$

  2. $a_1=a_2=a_3=...=a_n \ne 0$

  3. $a_1+a_2+...+a_n=0,$ $ a_1 \ne a_2$

  4. $a_1+a_2+...+a_n \ne 0$ , $a_1 \ne a_2$

  • 3
    Note that there isn't a simple answer in terms of $n$; it really depends what $v=(a_1,\ldots,a_n)$ is. For example, in $\mathbb{C}^3$, the dimensions given by $v=(0,0,0), (1,1,1), (1,0,-1), (1,2,3)$ are 0, 1, 2, and 3 respectively. Note that for both of these last two values of $v$, there are 6 distinct permutations of $v$, but the dimensions of the corresponding subspaces are different.2011-10-24
  • 0
    ok, what if the $a_1+a_2+a_3+...+a_n=0$ and $a_1 \ne a_2$?2011-10-24
  • 0
    Then you can reduce your problem to the case $(a_1,...,a_n)=(1,-1,0,0,...,0)$. Given any $(b_1,b_2,...,b_n)$ such that $b_1+b_2 + ... b_n = 0$ is a linear combination of these, so $(1,-1,0,...,0)$ generates an $n-1$ dimensional space.2011-10-24

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