How to know if a given (infinite) set of vectors lie in the same hyperplane?

Question

Fix $n \in \mathbb{N}$. Suppose $\alpha \in [0,1]$. Let $V_{\alpha}$ be the $n$-dimensional vector whose $k$-th component is given by $\alpha^{(k-1)}(1-\alpha)^{(n-k)},\; k \in [n]$. Now let $X$ be the set of all such vectors, i.e. $X = \{V_{\alpha}: \alpha \in [0,1]\}$. Can $X$ lie in the same hyperplane passing through the origin, i.e. does there exist any $n$-dimensional vector $A$ such that $A^TV_{\alpha}=0$ for all $V_{\alpha} \in X$?

Edit: With reference to $A.\Gamma.$'s answer, what if $\alpha \in (0,1)$?

Maybe this question is straightforward and I'm probably failing to understand something basic.

Answer 1

I assume that $a^{(k)}=a^k$.

1. There is no hyperplane that goes through the origin and contains all $V_\alpha$. To see that, set $\alpha=0$ and $\alpha=1$ to conclude that the first and the last coordinates of $A$ are zeros. Then cancel $\alpha(1-\alpha)$ and continue in the same way to set $\alpha=0,1$. Hence, $A=0$. Formally, this step is proved by induction.
2. There is a hyperplane that contains all $V_\alpha$: $$ 1=(\alpha+1-\alpha)^{n-1}=\sum_{k=1}^n\binom{n-1}{k-1}\alpha^{k-1}(1-\alpha)^{n-k}. $$