Difference of Projections

Theorem
Let $H$ be a Hilbert space.

Let $P, Q$ be projections.

Then the following are equivalent:


 * $(1): \quad P - Q$ is a projection
 * $(2): \quad P Q = Q$
 * $(3): \quad Q P = Q$

Proof
First it is shown that $(2)$ is equivalent to $(3)$.

Then, equivalence to $(1)$ is shown.

$(2)$ iff $(3)$
Suppose $P Q = Q$.

Then $P = P + Q - P Q$, and by Product of Projections, necessarily $P Q = Q P$.

That is, $Q P = P Q = Q$.

The converse is obtained by swapping the rôles of $P$ and $Q$ in Product of Projections.

$(2), (3)$ imply $(1)$
Suppose $P Q = Q P = Q$. Then as $P, Q$ are projections:


 * $\paren {P - Q}^2 = P^2 + Q^2 - P Q - Q P = P - Q$

That is, $P - Q$ is idempotent.

From Adjoining is Linear, $\paren {P - Q}^* = P^* - Q^* = P - Q$.

An application of Characterization of Projections, statement $(4)$ shows that $P - Q$ is a projection.

$(1)$ implies $(2)$
Let $P - Q$ be a projection.

Then by Characterization of Projections, statement $(6)$, one has:


 * $\innerprod {P h} h_H - \innerprod {Q h}, h_H \ge 0$

Applying this statement on $P, Q$ also, one obtains that:
 * $P h = 0 \implies Q h = 0$

that is:
 * $\ker P \subseteq \ker Q$

Orthocomplement Reverses Subset and that $P, Q$ are projections combine to state $\Rng Q \subseteq \Rng P$.

So for every $h \in H$, there is a $p \in H$ with $Q h = P p$.

It follows that:
 * $P Q h = P P p = P p = Q h$

Hence $PQ = Q$.

Also see

 * Sum of Projections
 * Product of Projections