Projections: Ordering

Question

Given a unital C*-algebra $1\in\mathcal{A}$.

Consider projections: $$P^2=P=P^*\quad P'^2=P'=P'^*$$

Order them by: $$P\leq P':\iff\sigma(\Delta P)\geq0\quad(\Delta P:=P'-P)$$

Then equivalently: $$P\leq P'\iff P=PP'=P'P\iff\Delta P^2=\Delta P=\Delta P^*$$

How can I check this?

(Operator algebraic proof?)

Answer 1

The assertion $P\leq Q$ means $Q-P\geq0$. Then $$ 0\leq P(Q-P)P=PQP-P\leq P^2-P=P-P=0. $$ So $P=PQP$. Now we can write this equality as $$0=P-PQP=P(I-Q)P=[(I-Q)P]^*[(I-Q)P],$$ so $(I-Q)P=0$, i.e. $P=QP$. Taking adjoints, $P=PQ$.

The converse also holds: if $P=PQ=QP $, then $$Q-P=Q^2-QPQ=Q (I-P)Q\geq0. $$

Edit: for the equivalence $Q-P\geq0$ iff $Q-P$ is a projection:

I think it is easier to show that $Q-P$ is a projection iff $P=PQ=QP$ (the latter being equivalent to $Q-P\geq0$ by the above).

If $Q-P$ is a projection, then $$Q-P=(Q-P)^2=Q+P-PQ-QP,$$ so $$\tag{1}2P=PQ+QP.$$ Multiplying by $I-P$ on the left, we get $(I-P)QP=0$, or $QP=PQP$. Taking adjoints, we obtain $QP=PQ$. Now $(1)$ is $2P=2PQ$, i.e. $P=PQ=QP$.

Conversely, if $P=QP=PQ$, then $$ (Q-P)^2=Q+P-PQ-QP=Q+P-2P=Q-P. $$

Answer 2

If $Q \ge P$, then \begin{align} \|Px\|^{2} & =\|QPx\|^{2}+\|(I-Q)Px\|^{2} \\ & = (PQPx,x)+\|(I-Q)Px\|^{2} \\ & \ge (PPPx,x)+\|(I-Q)Px\|^{2} \\ & = \|Px\|^{2}+\|(I-Q)Px\|^{2} \\ \implies & (I-Q)Px=0. \end{align}