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

$(2): \quad P Q = Q$

$(3): \quad Q P = Q$

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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.

$\Box$

$(2), (3)$ 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.

$\Box$

$(1)$ 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 $PQ = Q$.

$\blacksquare$

Also see

• Sum of Projections
• Product of Projections

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next) $II.3 \text{ Exercise } 6$