Show that the maximum size of a one-distance set in $\Bbb R^n$ is $n+1$ It is pretty believable that the maximum size of a one-distance set in $\Bbb R^n$ is $n+1$ (take the vertices of a simplicial complex). I am trying to prove this fact using linear algebra and haven't been able to formalize the result.
Let $\{v_0,v_1,\ldots,v_t\}$ be a one-distance set in $\Bbb R^n$. Without loss of generality $v_0=\vec 0$, so define the function
$$w_i=v_i-v_0=v_i$$
for $1\le i\le t$. Then note that the inner product $w_i\cdot w_j$ gives
\begin{align}
w_i\cdot w_j&=1&\text{if }i=j;\\
w_i\cdot w_j&=\delta<1&\text{if }i\neq j.
\end{align}
Now we aim to show that $w_i$ are linearly independent. Suppose
$$\sum_{i=1}^t\lambda_i w_i=\vec 0.$$
Then multiplication by $w_j$ gives something like 
$$\lambda_j+\delta(\lambda_1+\cdots+\lambda_{j-1}+\lambda_{j+1}+\cdots+\lambda_t)=0$$
and hopefully this implies $\lambda_j=0$. Thus, $w_i$ are linearly independent so $t\le n$, so the size of the original set is at most $t+1$.
Any suggestions for the last part?
 A: There's no hope for this proof to work as it stands, for suppose that you take a whole bunch of  (e.g., $2n$) distinct unit vectors $u_i$ in the first quadrant. Then you'll have
\begin{align}
u_i \cdot u_i &= 1 & \text{for all $i$, because these are unit vectors}\\
0 < u_i \cdot u_j &< 1 \text{for $i \ne j$, by Cauchy Schwartz.}
\end{align}
Since these are exactly the things you know about the $w_i$, you cannot conclude from these mere things that there are no more than $n$ of the $w_i$, or that the $w_i$ are independent. 
An inductive proof might be more successful. Use, as your hypothesis, that for $k$ points in a distance-1 set in $\mathbb R^n$, with the first point at the origin, there's an orthonormal basis in which the remaining $k$ points are the vertices of a standard $(k-1)$-dimensional simplex in the span of the first $k$ basis vectors, and $k < n+1$. 
(You'll need to pick your "standard" simplex; it may be simpler to work with affine coordinate frames rather that vector bases.) 
