# 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?

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 $P_iP_j=0$.

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 $j \ne k$.

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

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, $p_{i} p_{j} = 0_{A}$.