I am reading a paper that incidentally uses a bit of theory of symmetric tensor spaces. I came across the following claim:
If we're given linearly independent vectors
$x_1, \ldots, x_n \in \mathbb{R}^n$, then the tensors from the following set
$\{H_T = \sum_{\sigma \in S_{n-t}} \prod_{i \notin T} x_{T(\sigma(i))}:T \subset \{1,\ldots,n \}, |T|=t \}$
are linearly independent.
By $T(i)$ i mean the i-th element of the set $\{1, \ldots,n \} \; \backslash\; T$, and the product sign denotes tensor product.
I'm not really fluent in tensor algebras and when I tried proving it directly, expressing each $x_i$ in terms of standard basis, I drowned in the summing indices.
Any help regarding a proof of this claim would be appreciated.
Best Answer
The determinant of the $H_T$'s is a power of the determinant of the $x_i$'s, possibly times a nonzero numerical constant. See https://mathoverflow.net/questions/40077/determinant-and-symmetric-power where this is discussed.