Linear Independence of a set of $n!$ vectors

Question

Let $\{v_1,..,v_n\}$ be a linearly independent set of unit vectors in a finite dimensional real vector space. I want to show the set $\{v_{\sigma(1)}\otimes..\otimes v_{\sigma(n)}\}_{\sigma \in S_n}$ is also linearly independent, where $\otimes$ is the Kronecker product and $S_n$ is the symmetric group.

For $n=2$, it is clear that $v_1\otimes v_2$ and $v_2\otimes v_1$ are linearly independent as they are distinct unit vectors. I cannot understand how to proceed further. Any help will be appreciated.

Answer 1

Since you proved the case $n=2$, you can try an inductive approach. Here's a sketch.

First show the following lemma (which also implies the $n=2$ case directly):

Let $k,m \in \Bbb{N}$, let $x_1,\ldots, x_m \in \overbrace{V\otimes \cdots \otimes V}^{k}$ be arbitrary and let $y_1,\ldots, y_m \in V$ be linearly independent. Suppose that $$x_1\otimes y_1 + \cdots x_m \otimes y_m = 0.$$ Then $x_1 = \cdots = x_m = 0$.

Suppose the claim is true for all linearly independent sets of $n-1$ vectors. Assume now that $$ 0 = \sum_{\sigma \in S_n} \alpha_{\sigma} (v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)})$$ and we wish to prove that all scalars $\alpha_\sigma = 0$ for $\sigma \in S_n$. We have \begin{align} 0 &= \sum_{\sigma\in S_n} \alpha_{\sigma} (v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n)})= \sum_{j=1}^n \left(\sum_{\substack{\sigma\in S_n, \\ \sigma(n)=j}} \alpha_{\sigma} (v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n-1)})\right) \otimes v_j \end{align} so the lemma implies that $$\sum_{\substack{\sigma\in S_n, \\ \sigma(n)=j}} \alpha_{\sigma} (v_{\sigma(1)}\otimes \cdots \otimes v_{\sigma(n-1)})=0.$$ For a fixed $j \in \{1,\ldots, n\}$, the Kronecker product of vectors $v_1,\ldots, v_{j-1},v_{j+1}, \ldots, v_n$ in every possible ordering appears in the above sum exactly once. Therefore by the inductive hypothesis we have $\alpha_\sigma = 0 $ for all $\sigma \in S_n$ such that $\sigma(n)=j$. Since $j$ was arbitrary, the claim follows.