Sum of Projections/Binary Case

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Theorem

Let $H$ be a Hilbert space.

Let $P, Q$ be projections.


Then $P + Q$ is a projection if and only if $\Rng P \perp \Rng Q$.

Here, $\Rng P$ denotes range, and $\perp$ denotes orthogonality.


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Proof

Necessary Condition

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

\(\ds P + Q\) \(=\) \(\ds \paren {P + Q}^2\) $P + Q$ is an idempotent
\(\ds \) \(=\) \(\ds P^2 + P Q + Q P + Q^2\)
\(\ds \) \(=\) \(\ds P + P Q + Q P + Q\) $P$ and $Q$ are idempotents
\(\ds \leadstoandfrom \ \ \) \(\ds PQ + QP\) \(=\) \(\ds 0\)


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.

$\Box$


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:

\(\ds \paren {P + Q}^2\) \(=\) \(\ds P^2 + P Q + Q P = Q^2\)
\(\ds \) \(=\) \(\ds P + Q\) $P$ and $Q$ are idempotents


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.

$\blacksquare$


Also see


Sources

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