Projections: Orthogonality Given a unital C*-algebra $1\in\mathcal{A}$.
Consider projections:
$$P^2=P=P^*\quad P'^2=P'=P'^*$$
Order them by:
$$P\perp P':\iff\sigma(\Sigma P)\leq1\quad(\Sigma P:=P+P')$$

Then equivalently:
  $$P\perp P'\iff 0=PP'=P'P\iff\Sigma P^2=\Sigma P=\Sigma P^*$$

How can I check this?
(Operator algebraic proof?)

Let's enumerate the conditions:
(1) $P\perp P'$;
(2) $0=PP'=P'P$;
(3) $\Sigma P$ is a projection.
We can assume that $A\subset B(H)$ for some Hilbert space $H$ (this basically follows from the GNS construction); also, remember that for positive operators $T\in B(H)$ we have $\langle T\xi,\xi\rangle\geq 0$ for all $\xi\in H$.
For (1)$\Rightarrow$(2), assume that $\xi\in P(H)$ and note that $\langle (P+P')\xi,\xi\rangle\leq\langle\xi,\xi\rangle=\|\xi\|^2$ by assumption (since $0\leq P+P'\leq 1$). Since $\langle P\xi,\xi\rangle=\|\xi\|^2$, we have $ \|P'\xi\|^2=\langle P'\xi,\xi\rangle=0$, so that $\xi=(1-P')\xi+P'\xi=(1-P')\xi$. Thus for all $\eta\in H$, note that $P\eta\in P(H)$, so that we have $P\eta=(1-P')P\eta$ by what we found above. This means that $P'P\eta=0$ for all $\eta\in H$, so $P'P=0$. Taking the adjoint yields (2). (3) follows from (2) by just computing, and any projection $Q\in A$ satisfies $0\leq Q\leq 1$, so if (3) holds, then $\sigma(\Sigma P)\subset[0,1]$ immediately.
