Sum of Projections/Binary Case

Theorem
Let $H$ be a Hilbert space.

Let $P, Q$ be projections.

Then $P + Q$ is a projection $\Rng P \perp \Rng Q$. Here, $Rng P$ denotes range, and $\perp$ denotes orthogonality.

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

Now suppose that $h \in \Rng Q$; say $h = Q q$ for $q \in H$.

Then it follows that $Q h = Q Q q = Q q = h$ as $Q$ is idempotent.

It follows that:
 * $0 = P Q h + Q P h = P h + Q P h$

From Characterization of Projections, statement $(6)$, $\innerprod {Q P h} {P h}_H \ge 0$.

Next, observe:
 * $0 = \innerprod {P h + Q p h} {P h}_H = \innerprod {P h} {P h}_H + \innerprod {Q P h} {P h}_H$

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

Hence:
 * $h \in \Ker P = \paren {\Rng P}^\perp$

It follows that $\Rng Q \perp \Rng P$, as asserted.

Sufficient Condition
Suppose that $\Rng P \perp \Rng Q$.

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


 * $\Rng P \subseteq \Ker Q$
 * $\Rng Q \subseteq \Ker P$

That is, for all $h \in H$:
 * $Q P h = P Q h = \mathbf 0_H$

Hence:

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

Furthermore, by Adjoining is Linear, have:


 * $\paren {P + Q}^* = P^* + Q^* = P + Q$

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