Sum of projections onto vectors which sum to zero. Premise: Let $V$ be some vector space over $K$ (in my case it's $d$ dimensional Euclidian space). Let $U= \{\mathbf u_i:i=1...n\}$ be a subset of $V$ which has the additional properties $\sum_i \mathbf u_i = 0_V$ and $|\mathbf u_i|=1_K \,\,\forall i$. 
The projection operator $P :  V \rightarrow V$ onto some vector $\mathbf u_i \in V$ is defined $P(\mathbf v) = (\mathbf u_i \otimes \mathbf u_i) \cdot \mathbf v$. 
Consider the quantity $\left(\sum_i \mathbf u_i \otimes \mathbf u_i\right)\cdot\mathbf v = \sum_i \mathbf u_i (\mathbf u_i \cdot\mathbf v) := P_{tot}(\mathbf v)$. 
Question: Is there nice simplification for this quantity? Is it the case that $P_{tot}(\mathbf v) = \lambda \mathbf v$ for some $\lambda \in K$?
Edit: I realised I was barking up the wrong tree and that in general $P_{tot}(\mathbf v)\neq 0_V$
 A: I'll use a slightly less abstract notation to express the answer. The projection onto a unit-vector $u_i$ is given by $Proj_{u_i}(v) = (u_i \cdot v)u_i$ where $u_i \cdot v$ denotes the dot-product of $u_i$ and $v$. Geometrically, this formula is clear; it selects the $u_i$ component then multiplies by $u_i$ to create a vector with the length of the $u_i$ component in the $u_i$-direction. It is the $u_i$-vector component of $v$. Terminology settled. Consider the given data:
$$ \sum_{i=1}^k u_i=u_1+ \cdots u_k=0  $$
We wish to show that $L(v) = \sum_{i=1}^k Proj_{u_i}(v)$ vanishes. Or perhaps, to see that it vanishes when $v \neq 0$ and $v \notin \text{span}\{u_1, \dots , u_k \}$. Very well, the mystery will soon be revealed, let us investigate: observe
$$ L(v) = \sum_{i=1}^k Proj_{u_i}(v) = \sum_{i=1}^k (u_i \cdot v)u_i $$
If we examine $v=u_j$ for a particular $j \in \{ 1,2,\dots , k\}$ then
$$ L(u_j) = \sum_{i=1}^k (u_i \cdot j)u_i = \sum_{i=1}^k \delta_{ij}u_i = u_j $$
We have linear dependence of $\{ u_1, \dots u_k\}$ by assumption. Reorder them if necessary and relabel so that the first $r <k$ vectors are LI; $\beta = \{ u_1, \dots , u_r \}$ forms an orthonormal basis for $W = \text{span} \{ u_1, \dots , u_k \}$. Finally, extend $\beta$ to $\beta \cup \{ w_1, \dots, w_{n-r} \}=\gamma$ by adjoining vectors outside $W$. By the gram-schmidt algorithm we may take $\gamma$ to be orthonormal. Therefore, we know $w_i \cdot u_j=0$ and $w_i \cdot w_j = \delta_{ij}$. Let $v = x+y$ where $x \in W$ and $y \in W^{\perp} = \text{span}\{ w_1, \dots ,w_{n-r}\}$. Return to our calculation of $L$ with all this in mind,
$$ L(v) = L(x+y) = L(x)+L(y) $$
However,
$$ L(x) = L( \sum_{i=1}^r x_iu_i) = \sum_{i=1}^r x_iL( u_i)=\sum_{i=1}^r x_iu_i=x. $$
whereas,
$$ L(y) = L( \sum_{j=1}^{n-r} y_jw_j) = \sum_{j=1}^{n-r} y_j \sum_{i=1}^r(u_i \cdot w_j)u_i=\sum_{j=1}^{n-r} y_j \sum_{i=1}^r(0)u_i=0. $$
Therefore, $L|_W = Id_W$ whereas $L|_{W^{\perp}}=0$. Notice that $P \neq L$ since the $P$ has all the redundant vectors built into the formula. In our current notation, we can write
$$ P(v) = \sum_{i=1}^k (u_i \cdot v)u_i=\sum_{i=1}^r (u_i \cdot v)u_i+\sum_{i=r+1}^k (u_i \cdot v)u_i=L(v) + \sum_{i=r+1}^k (u_i \cdot v)u_i$$
If $v \notin W$ then $P(v)=0$ is clear. However, if $v \in W$ then $v = \sum_{j=1}^k v_j u_j$ and $L(v)=v$ hence:
\begin{align} 
P(v) &= v+\sum_{i=r+1}^k \left(u_i \cdot \left[\sum_{j=1}^{k}v_ju_j \right] \right)u_i \\
&= v+\sum_{i=r+1}^k \sum_{j=1}^{k}v_j \left(u_i \cdot u_j  \right)u_i \\
&= v+\sum_{i=r+1}^k \sum_{j=1}^{k}v_j \delta_{ij}  u_i \\
&= v+\sum_{i=r+1}^k v_i  u_i \\
&= \sum_{i=1}^r v_i  u_i + 2\sum_{i=r+1}^k v_i  u_i
\end{align}
I don't believe this is zero. We could use $u_1 = -\sum_{j=2}^k u_j$ to eliminate $u_1$ from the sum etc... but nothing particularly pretty jumps out at me. So, sorry, not zero.
A: It is not in general true given the conditions above that $P_{tot}(\mathbf v) = \lambda \mathbf v$.
Proof: 
Let  $\dim(V) = d$ and $U^\prime = \{\mathbf u_1,...\mathbf u_d\}$ be a basis for V. Then consider the set $U = \{\mathbf u_1,...\mathbf u_d, -\mathbf u_1, ... -\mathbf u_d\}$; it satisfies the properties $\sum_i \mathbf u_i = 0_v$ as well as $|\mathbf u_i| = 1 \,\,\forall i$.
In general our vectors $U^\prime$ need not be orthonormal. Consider the case where we have $\mathbf u_1 \cdot \mathbf u_2 \neq 0$ and $\mathbf u_1 \cdot \mathbf u_i = 0$ for $i=3,...d$.
Let $V^\prime \subset V$ be the subspace of $V$ such that $\forall \,\, \mathbf v \in V^\prime $ we have $\mathbf v \cdot \mathbf u_1 = 0$. The operator $P_{tot}$ applied to elements of this subspace is
$$P_{tot}(\mathbf v) = 2\sum_{i=2}^d (\mathbf u_i \cdot \mathbf v) \mathbf u_i.$$
because $\mathbf u_1\cdot\mathbf v=0$ and $(\mathbf u_{i+d}\cdot\mathbf v)\mathbf u_{i+d} = (-\mathbf u_{i}\cdot\mathbf v)(-\mathbf u_{i}) = (\mathbf u_{i}\cdot\mathbf v)\mathbf u_{i}$ if $d<i\leq n$. Clearly then $\text{span}(P_{tot}(V^\prime)) \subset \text{span}(V)$ [proper subset] since it includes no $\mathbf u_1$ component and $U^\prime$ was a basis for $V$.
Consider the vector $\mathbf w \in \text{span}(P_{tot}(V^\prime))$, $\mathbf w = \sum_{j=2}^d w^j\mathbf u_j \neq 0$. If $\mathbf w = \lambda \mathbf v$ we require $\mathbf w \cdot \mathbf u_1 = 0$, but notice
$$\mathbf w \cdot \mathbf u_1 = \sum_{j=2}^d w^j\mathbf u_j\cdot \mathbf u_1,$$
which from our construction of $U^\prime$ gives $\mathbf w \cdot \mathbf u_1 = w^2\mathbf u_1 \cdot \mathbf u_2 \neq 0$. 
Thus if $\mathbf w = P_{tot}(\mathbf v)$ it is not true in general that $\mathbf w = \lambda \mathbf v$.
Addendum:
The examples I was looking at on the computer were returning $P_{tot}(\mathbf v) = \frac{n}{2}\mathbf v$ for $\mathbb R^2$. It turns out the examples I constructed all had some additional symmetries which made this be the case.
