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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 $i \neq j$?

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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 $$ p_ip_j=0\qquad (\forall 1\leq i\ne j\leq n). $$

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  • $\begingroup$ Thank you so much for that marvellous answer. Najlepša hvala za odličen odgovor! :-) $\endgroup$
    – Lisa
    Jan 24 '15 at 16:26
  • $\begingroup$ @Lisa You speak slovene!!! Torej lahko v bodoče pišem v slovenščini :-) $\endgroup$ Jan 24 '15 at 17:12
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    $\begingroup$ This should probably be clear, but I still don't see why we have $\langle (p_ip_j + p_jp_i)\xi,\xi\rangle \geq 0$. $\endgroup$
    – Lisa
    Jan 24 '15 at 18:02
  • $\begingroup$ @Lisa You are right, there could be a gap. I will have to rethink my answer. $\endgroup$ Jan 24 '15 at 18:22
  • $\begingroup$ @Lisa I have added an answer to your question. The idea of the solution is from math.stackexchange.com/questions/1119568/…, so please upvote the solution there. Hvala :-) $\endgroup$ Jan 26 '15 at 5:51

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