Sum of Projections/Binary Case

Theorem
Let $H$ be a Hilbert space.

Let $P, Q$ be projections.

Then $P + Q$ is a projection $\operatorname{ran} P \perp \operatorname{ran} Q$. Here, $\operatorname{ran}$ denotes range, and $\perp$ denotes orthogonality.

Necessary Condition
Suppose $P + Q$ is a projection. Then:

Now suppose that $h \in \operatorname{ran} Q$; say $h = Qq$ for $q \in H$.

Then it follows that $Qh = QQq = Qq = h$ as $Q$ is idempotent.

It follows that $0 = PQh + QPh = Ph + QPh$.

From Characterization of Projections, statement $(6)$, $\left\langle{QPh, Ph}\right\rangle_H \ge 0$.

Next, observe $0 = \left\langle{Ph + Qph, Ph}\right\rangle_H = \left\langle{Ph, Ph}\right\rangle_H + \left\langle{QPh, Ph}\right\rangle_H$.

As the second term is non-negative, the first is non-positive; it follows that $Ph = \mathbf{0}_H$ from the definition of the inner product.

Hence $h \in \operatorname{ker} P = \left({\operatorname{ran} P}\right)^\perp$.

It follows that $\operatorname{ran} Q \perp \operatorname{ran} P$, as asserted.

Sufficient Condition
Suppose that $\operatorname{ran} P \perp \operatorname{ran} Q$.

Then as $P, Q$ are projections, have:


 * $\operatorname{ran} P \subseteq \operatorname{ker} Q$
 * $\operatorname{ran} Q \subseteq \operatorname{ker} P$

That is, for all $h \in H$, one has $QPh = PQh = \mathbf{0}_H$. Hence:

That is, $P + Q$ is an idempotent.

Furthermore, by Adjoining is Linear, have:


 * $\left({P + Q}\right)^* = P^* + Q^* = P + Q$

where the latter follows from Characterization of Projections, statement $(4)$.

This same statement implies that $P+Q$ is also a projection.

Also see

 * Product of Projections
 * Difference of Projections