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As long as the $n$ entries of the vector are all different and they dont add up to zero. If it is true, how to prove it, if not, what is a counter example?

DeepSea
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Rikka
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  • It is a hardship for Readers to put the problem statement in the title and omit it from the body of the Question. Complete sentences in stating the problem are a help. – hardmath Jul 27 '16 at 20:47
  • As I understand it, the vector $(1,1,1,1.....,1)$ is a clear counterexample. Please elaborate on your posing of the question. – Zestylemonzi Jul 27 '16 at 20:52
  • ooops sorry! Thanks for pointing that out. – Zestylemonzi Jul 27 '16 at 20:53
  • sorry that i put the requirements of the vector on the body – Rikka Jul 27 '16 at 20:54
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    Is $R$ = the real numbers? If so, then yes. The claim is equivalent to saying that the defining representation of the symmetric group $S_n$ has only two nontrivial proper subrepresentations: the one formed by all multiples of the vector $\left(1,1,\ldots,1\right)$, and the one formed by all vectors whose entries sum to $0$. This should be in all kinds of sources (e.g., Fulton-Harris). – darij grinberg Jul 27 '16 at 21:11
  • Is there any 'elementary proof'? sorry that i have not learnt abstract algebra yet. – Rikka Jul 27 '16 at 21:29
  • Hint: It suffices to show that the permutations of the vector (let me call it $v$) span $R^n$. By assumption, $v$ has two distinct entries. By switching them, you get a permutation $v'$ of $v$. Now, the vector $v'-v$ is a nonzero multiple of the vector $e_i-e_j$ for two distinct indices $i$ and $j$ (where $e_k$ means the $k$-th basis vector of the standard basis of $R^n$). Thus, $e_i-e_j$ lies in the span of the permutations of $v$. But since you can also permute the entries of $v$ at will, you can show that $e_i-e_j$ lies in the span of the permutations of $v$ for every ... – darij grinberg Jul 27 '16 at 22:23
  • ... two distinct indices $i$ and $j$. These $e_i - e_j$, together, span the subspace of $R^n$ formed by all vectors whose coordinates sum to $0$. So all such vectors lie in the span of the permutations of $v$. Moreover, by adding multiples of $v$, you can get every vector in $R^n$ (since the coordinates of $v$ do not sum to $0$). This is the gist of the argument. The details are somewhat painful to write up. – darij grinberg Jul 27 '16 at 22:24
  • Actually, there's a slicker argument at http://math.stackexchange.com/questions/75427/subspace-generated-by-permutations-of-a-vector-in-a-vector-space , which works for any field instead of $R$. – darij grinberg Jul 27 '16 at 22:27
  • Wow, what a beautiful proof, i am really too stupid to attack the question on that angle. I do love math, however I had already encountered a lot of difficult problems for the first year of study, and I am a bit discouraged. Do you have any advice on what to do if I REALLY want to pursue mathematics? – Rikka Jul 28 '16 at 02:48
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    @Rikka Yes: remember that it's okay to feel stupid. That's bound to happen over and over if you pursue mathematics; ask any mathematician. – Ben Grossmann Jul 28 '16 at 12:20

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