TITLE: Orthogonal projections with $\sum P_i =I$, proving that $i\ne j \Rightarrow  P_{j}P_{i}=0$
QUESTION [16 upvotes]: 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?

REPLY [3 votes]: 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
$$\left(\sum_k P_k\right)^2=\sum_k P_k + \sum_{i<j}(P_i P_j + P_j P_i)=\sum_k P_k.$$