Prove that in a C*-algebra only orthogonal projections can sum to a projection Let $A$ be a $C^*$-algebra, and let $p_1, \ldots, p_n \in A$ be projections, meaning $p_i = p_i^* = p_i^2$. Now assume that the sum $p = p_1 + \ldots + p_n$ is also a projection. How can one show that this implies that the $p_i$'s must be orthogonal, i.e. $p_ip_j = 0$ whenever $i \neq j$?
 A: Since
$$ p=p^2=(p_1+\cdots+p_n)^2=\sum_{i=1}^{n}\sum_{j=1}^{n}p_i p_j= $$
$$=p_{1}^{2}+\cdots+p_{n}^{2}+\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)=p+\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)$$
we have
$$ \sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i)=0. \tag5 $$
Without loss of generality we may assume that $A$ is a $C^*$-subalgebra of $B(H)$, where
$H$ is a suitable complex Hilbert space. For any $\xi \in H$, we have
$$ \langle (p_ip_j+p_jp_i)\xi,\xi \rangle \geq 0. \tag6 $$
It follows from
$$ 0=\langle (\sum_{1\leq i<j\leq n}(p_ip_j+p_j p_i))\xi,\xi\rangle=\sum_{1\leq i<j\leq n}\langle (p_ip_j+p_j p_i)\xi,\xi\rangle \qquad \forall \xi \in H.$$
that
$$ \langle (p_ip_j+p_j p_i)\xi,\xi\rangle=0\qquad \forall \xi \in H, $$
i.e., 
$$ p_ip_j+p_j p_i=0 \tag1. $$
It follows from (1) that
$$ p_ip_j+p_j p_ip_j=0 \tag2$$
and
$$ (1-p_j)(p_ip_j+p_j p_i)=0. \tag3 $$
Since $(1-p_j)p_jp_i=0$ it follows from (3) that
$$ p_ip_j-p_j p_ip_j=0. \tag4$$
We conclude from (2) and (4) that $p_ip_j=0$.
Appendix
Unfortunately the above reasoning is not correct if $n>2$ because (6) does not hold in general. For $n=2$, (6) holds because of (5). However, for $n\geq 3$ I cannot see how to deduce (1) from (5).
Let us show that (6) does not hold in general. Let $p, q$ be orthogonal projections on a Hilbert space $H$. For any $\xi H$, one has
$$ \langle (pq+qp)\xi,\xi\rangle=\langle pq\xi,\xi\rangle+\langle qp\xi,\xi\rangle=\langle q\xi,p\xi\rangle+\langle p\xi,q\xi\rangle=2\Re(\langle q\xi,p\xi\rangle)=2\Re(\langle pq\xi,\xi\rangle).   $$
Hence $ \langle (pq+qp)\xi,\xi\rangle\geq 0$ for any $\xi \in H$ if and only if 
$\Re(\langle pq\xi,\xi\rangle)\geq 0$ for any $\xi \in H$. However, as the following example shows, this last inequality does not hold for any pair of orthogonal projections.
Example
Let $H={\mathbb C}^*$ and $A={\mathbb M}_2$. Let
$$p=\left( \begin{array}{cc} \frac{1}{2} & \frac{1}{2}\\ \frac{1}{2} & \frac{1}{2}\end{array} \right)\quad \text{and}\quad q=\left( \begin{array}{cc} 1 & 0\\ 0 & 0 \end{array}\right). $$
It is easy to check that $p^2=p=p^*$ and $q^2=q=q^*$. Let $\xi=\left(\begin{array}{r}1\\-2\end{array}\right)$. Then $ \langle pq\xi,\xi\rangle =-\frac{1}{2}$, i.e., $ \Re( \langle pq\xi,\xi\rangle)<0$ and therefore, by the above observation, $\langle (pq+qp)\xi,\xi\rangle<0$.
Answer
Since $p=p_1+\cdots+p_n$ is an orthogonal projection its norm is $1$. Hence, for any $\xi \in H$, one has
$$ |\langle (p_1+\cdots+p_n)\xi,\xi\rangle|=|\langle p\xi,\xi\rangle|\leq \| p\| \|\xi \|^2=\|\xi\|^2. $$
On the other hand,
$$ |\langle (p_1+\cdots+p_n)\xi,\xi\rangle|=|\langle p_1\xi,\xi\rangle+\cdots+\langle p_n\xi,\xi\rangle|=|\langle p_1\xi,p_1\xi\rangle+\cdots+\langle p_n\xi,p_n\xi\rangle|= $$
$$=\| p_1 \xi\|^2+\cdots+\| p_n \xi\|^2. $$
Hence
$$\| p_1 \xi\|^2+\cdots+\| p_n \xi\|^2\leq \| \xi\|^2\quad (\forall \xi \in H). \tag7$$
Put $p_j\xi$ in (7) istead of $\xi$, then
$$\| p_1p_j \xi\|^2+\cdots+\| p_j \xi\|^2+\cdots+\| p_n \xi\|^2\leq \| p_j\xi\|^2\quad (\forall \xi \in H) $$
which gives
$$\| p_1p_j \xi\|^2+\cdots+\| p_{j-1}p_j \xi\|^2+\| p_{j+1}p_j \xi\|^2+\cdots+\| p_n p_j\xi\|^2=0 \quad (\forall \xi \in H), $$
that is
$$ p_ip_j=0\qquad (\forall 1\leq i\ne j\leq n). $$
