I've seen the following statement (in some form or another) in many books. I'm wondering why it's true, and am trying to prove it:
If $[x_1], \cdots, [x_k] \in \mathbb{P}V$ are distinct, and not all the $[x_j]$ are colinear, then $x_1^d, \cdots, x_k^d$ are linearly independent for all $d \geq k-2$.
The $x_j^d$ are vectors in $S^d V$, the space of symmetric, order $d$ tensors over $V$. Specifically, $x_j^d = x_j \otimes \cdots \otimes x_j$. But, they can simply be thought of as polynomials of degree $d$.
My attempt at a proof of this statement: Without loss of generality, say that $k = d + 2$ and that all points lie on some $\mathbb{P} A \subset \mathbb{P} V$. Then, by contradiction, suppose that the images of the points are contained in some hyperplane $H_\alpha \in S^d A$. Then, $\alpha \in S^d A^\ast$ is a homogeneous polynomial of degree $d$ in $\mathrm{dim}(A)$ variables.
I want to obtain some contradiction by arguing that $\alpha$ vanishes at more points than it should (we know that $\alpha$ vanishes at $x_1, \cdots x_k$), but I don't think there's a general statement I can make about the number of roots that $\alpha$ has in $\mathbb{P} A$.
How can I obtain a contradiction?
There's a simple proof, using some representation theory. First, assume that all of the vectors lie on a single line, except for one. That is, making some choice, $V = E \oplus W$, where $x_1, \cdots, x_{k-1} \in E$, and $x_k \in W$.
Then,
$$S^d V = S^d (E \oplus W) = S^d E \oplus S^{d-1} E \otimes W \oplus \cdots \oplus S^d W$$
which is the only result we need from representation theory. It's pretty clear that $x_i \in S^d E$ for all $i = 1, \cdots, k-1$. And, $x_k \in S^d W$. But, clearly, $S^d E$ and $S^d W$ are disjoint (note that projective space doesn't include 0).
The general case follows easily from this.