Wrapping my head around some Linear Algebra Consider the vector space $\Bbb R^d$, where $d$ is a positive integer.  
Suppose I have $d+1$ unit vectors $\{v_1,v_2,...,v_{d+1}\} \in \Bbb R^d$ satisfying $\langle v_i, v_j \rangle = -1/d$ for all $i \neq j$.
I know that this condition guarantees that the vectors are "symmetric" around the origin. 
(How can I prove that their sum is zero?)
Let $W \in \Bbb R^d$.  Consider the expression $\sum_{i=1}^d v_i \langle v_i, W \rangle $.
In some sense, because $\{v_1,v_2,...,v_{d+1}\}$ is not an orthonormal set,the above expression gathers too many parts of $W$.  
How can I adjust this expression (not the unit vectors) so that it is equal to $W$?
I have a hunch it could be: $$W=\frac{d}{d+1} \sum_{i=1}^d  v_i \langle v_i, W \rangle $$
... but don't take my word on it.
Thank you very much for your help!
 A: To show their sum is $0$, I think this should work. Notice that for $\mathbb R^2$, we get $<v_1+v_2+
v_3, v_1+v_2, v_3>=(1-1/2-1/2+1)+(1-1/2-1/2)+(1-1/2-1/2)$. For each $v_i$, we get one $1$ in the expansion (since the $v_i$ are unit vectors, we get $<v_i,v_i>=1$, and we get $d$ copies of $-1/d$, for a total sum of $d(1-d(1/d)=0)$): $$<v_1+...+v_d, v_1+...+v_d>=d(1-(d)1)/d)=0 $$. Then $v_1+v_2+....+v_3$ is orthogonal to itself, so it must be the $0$ vector.
For the second question, i.e. (if I understood you correctly), if {$v_i$} is a generating set (it will not be a basis, since it has too-many vectors to be linearly-independent.), you can always use Gram-Schmidt algorithm to turn it into an orthonormal set. Then any vector $w$ in $W$ has an expansion: $$\Sigma_{i=1,2,..,n}<w,v_i>v_i $$ . I hope I understood your question correctly.  
A: I'm going to assume that we know that $\sum\limits_{i=1}^{d+1} v_i = 0$ and that $\operatorname{span}\{v_i : 1 \leq i \leq d+1\} = \Bbb{R}^d$. Now, let $X \in \Bbb{R}^d$ and let $X^i \in \Bbb{R}$ such that $X = \sum\limits_{i=1}^{d+1}X^i v_i$. Now, let's consider the sum $\sum v_i \langle v_i, X \rangle$.
$$
\sum_{i=1}^{d+1} v_i \langle v_i, X \rangle = \sum_{i=1}^{d+1} \sum_{j=1}^{d+1} X^j v_i \langle v_i, v_j \rangle = - \frac{1}{d} \sum_{i\neq j} X^j v_i + \sum_{i=1}^{d+1} X^i v_i \\
= - \frac{1}{d} \sum_{j=1}^{d+1} X^j \underbrace{\sum_{i=1}^{d+1} v_i}_{=0} + \Big[1+ \frac{1}{d} \Big] \underbrace{\sum_{i=1}^{d+1} X^i v_i}_{=X} \\= \Big[1+ \frac{1}{d} \Big] X = \frac{d+1}{d} X
$$ 
therefore, as your initial hunch suggested
$$
\sum_{i=1}^{d+1} v_i \langle v_i, X \rangle  = \frac{d+1}{d} X \implies X = \frac{d}{d+1} \sum_{i=1}^{d+1} v_i \langle v_i, X \rangle.
$$
