Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow  P_{j}P_{i}=0$ I am reading Introduction to Quantum Computing by Kaye, Laflamme, and Mosca. As an exercise, they write
"Prove that if the operators $P_{i}$ satisfy $P_{i}^{*}=P_{i}$ and $P_{i}^{2}=P_{i}$ , then $P_{i}P_{j}=0$ for $i\ne j$.'' 
In the context of this problem, it has been assumed that $I=\sum_{i=1}^{n} P_{i}$, where I suppose that $n$ could be infinite. I have shown that this is true in the trivial case $n=2$, but the general case has been eluding me. How should I attack this?
 A: For all $i,j$ you have $P_i+P_j\le\sum_k P_k\le I$, hence $P_i\le I-P_j$.
Multiplying by $P_j$ on left and right on LHS and RHS you then get
$$P_j P_i P_j\le P_j(I-P_j)P_j=0,$$
hence $P_j P_i P_j=0$, which implies $P_j P_i P_j=(P_j P_i)(P_j P_i)^\dagger=0$ and thus $P_i P_j=P_j P_i=0$.

You can also prove the other direction: if $P_i P_j=0$ for all $i\neq j$ then $\sum_k P_k$ is a projector, as
$$\left(\sum_k P_k\right)^2=\sum_k P_k + \sum_{i<j}(P_i P_j + P_j P_i)=\sum_k P_k.$$
A: For each $j$,
$$P_j=P_jIP_j=P_j\left(\sum_{k=1}^n P_k\right)P_j=\sum_{k=1}^nP_jP_kP_j=P_j+\sum_{k\neq j}P_jP_kP_j,$$ so $\sum\limits_{k\neq j}P_jP_kP_j=0$.  For each $i\neq j$, $P_jP_iP_j=(P_iP_j)^*P_iP_j$ is a positive operator, and a sum of positive operators is positive, so $-P_jP_iP_j=\sum\limits_{k\neq i,j}P_jP_kP_j$ is also positive.  This is only possible if $P_jP_iP_j=0$.  Since $\|P_iP_j\|^2=\|(P_iP_j)^*P_iP_j\|=\|P_jP_iP_j\|$, it follows that $P_iP_j=0$.
A: Because of the properties you state, $\|P_{j}x\|^{2}=(x,P_{j}x)=(P_{j}x,x)$. Therefore,
$$
           \|x\|^{2} = (\sum_{j}P_{j}x,x)= \sum_{j}\|P_{j}x\|^{2}.
$$
Apply this identity to $x=P_{k}y$, and use the fact that $P_{k}^{2}=P_{k}$:
$$
               \|P_{k}y\|^{2} = \sum_{j\ne k}\|P_{j}P_{k}y\|^{2}+\|P_{k}y\|^{2}.
$$
The only way this can happen is $P_{j}P_{k}y=0$ for all $j \ne k$.
A: Here is a slight variant of Jonas’ argument.

Assume that $ p_{1},\ldots,p_{n} $ are projection elements of a unital $ C^{*} $-algebra $ A $, where $ n \in \Bbb{N}_{\geq 2} $, such that
$$
\sum_{k = 1}^{n} p_{k} = 1_{A}.
$$
Choose distinct $ i,j \in [n] $, where $ [n] \stackrel{\text{df}}{=} \Bbb{N}_{\leq n} $. Then
\begin{align}
    p_{i}
& = p_{i} 1_{A} \\
& = p_{i} \sum_{k \in [n]} p_{k} \\
& = \sum_{k \in [n]} p_{i} p_{k} \\
& = p_{i}^{2} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} \\
& = p_{i} + p_{i} p_{j} + \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k}.
\end{align}
It follows that
$$
  p_{i} p_{j} p_{j}^{*}
= p_{i} p_{j}
= - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k}
= - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*},
$$
and consequently,
$$
(\spadesuit) \qquad
  (p_{i} p_{j}) (p_{i} p_{j})^{*}
= p_{i} p_{j} p_{j}^{*} p_{i}^{*}
= - \sum_{k \in [n] \setminus \{ i,j \}} p_{i} p_{k} p_{k}^{*} p_{i}^{*}
= - \sum_{k \in [n] \setminus \{ i,j \}} (p_{i} p_{k}) (p_{i} p_{k})^{*}.
$$
On the extreme left of $ (\spadesuit) $, we have a positive element, while on the extreme right of $ (\spadesuit) $, we have a negative element. This can only mean that both extremes are zero, so $ (p_{i} p_{j}) (p_{i} p_{j})^{*} = 0_{A} $. Hence,
$$
  \| p_{i} p_{j} \|_{A}^{2}
= \| (p_{i} p_{j}) (p_{i} p_{j})^{*} \|_{A}
= \| 0_{A} \|_{A}
= 0,
$$
or equivalently, $ p_{i} p_{j} = 0_{A} $.
