Given a bundle of k linear independent vectors, each vector coordinates is some permutation of $\{1, \ldots, n\}$. Prove that linear span of these vectors contains not more than $k!$ vectors whose coordinates are permutation of $\{1,\ldots,n\}$.
My observations
- call k linear independent vectors $\sigma_1, \ldots, \sigma_k$. Then if $\sigma=a_1\cdot\sigma_1+\cdots+a_k\cdot\sigma_k$(*), then $a_1+\cdots+a_k=1$ (write $(*)$ as linear system, sum all n rows and divide for $(1+\cdots+n)$
- $k!$ is achievable. Let take $\sigma_1=\operatorname{id}$, $\sigma_i$ is $\operatorname{id}$ with cyclically shifted first $k$ coordinates ($k=3$: $(1,2,3,4,5) \rightarrow (3,1,2,4,5) \rightarrow (2,3,1,4,5)$ they are independent (easy to calculate determinant)). Let's take $k!$ permutations of $\{1,\ldots,k\}$ and just add $\{k+1,\ldots,n\}$ at the end. It is obvious that rank of this bundle is less than or equal $k$. (But we need it is at least $k$.)
I don't need full solution, but I really ask for hint. Problem is beautiful but i'm at a dead end. The only one idea is induction. But it just don't work, because different bundles of k linear independent vectors have so different spans.
