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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. $a_1+a_2+...+a_n \ne 0$ , $a1 \ne a_2$

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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. – Brad Oct 24 '11 at 16:16
ok, what if the $a_1+a_2+a_3+...+a_n=0$ and $a_1 \ne a_2$? – Mohan Oct 24 '11 at 16:23
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. – Thomas Andrews Oct 24 '11 at 16:56

Let $V$ be the subspace spanned by the permuted versions of the vector $a=(a_1,a_2, ... ,a_n) \in K^n$. Let $U$ be the complementary subspace w.r.t. the usual bilinear form $\langle x,y\rangle=\sum_{i=1}^nx_iy_i$ of $K^n$, i.e. $$ U=\{ x\in K^n\mid \langle x,y\rangle=0\;\text{for all $y\in V$}\} $$ Then it is impossible for both $V$ and $U$ to have a vector with non-equal components. Assume contrariwise that $a_i\neq a_j$ for some pair of indices $i<j$, and that there exists $u=(u_1,u_2,\ldots,u_n)\in U$ such that $u_k\neq u_\ell$ for some pair of indices $k<\ell$. Then we can find two permuted versions of $a$: one, call it $v$, where $a_i$ occurs at position $k$ and $a_j$ at position $\ell$, and another, call it $w$, gotten from $v$ by swapping those two entries. Then $$ 0=0-0=\langle u,v\rangle-\langle u,w\rangle=\langle u,v-w\rangle=(u_k-u_\ell)(a_i-a_j), $$ which is a contradiction.

The conclusion is that one of the subspaces $U$ or $V$ is of dimension at most 1. As they are complementary $\dim U +\dim V=n$, and we can deduce that the dimension of $V$ is either 0,1, $n-1$ or $n$. It is clear, when either of the first two cases occurs. If $\dim V=n-1$, then $\dim U=1$, and, by the above observation, $U$ is spanned by the all ones vector, so $V$ is the zero-sum subspace. When none of these cases applies, we must have $\dim V=n$.

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Nice.${}{}{}{}$ – Gerry Myerson Oct 26 '11 at 12:26

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