# Difference of Projections

## Contents

## Theorem

Let $H$ be a Hilbert space.

Let $P, Q$ be projections.

Then the following are equivalent:

- $(1): \qquad P - Q$ is a projection
- $(2): \qquad PQ = Q$
- $(3): \qquad QP = Q$

## Proof

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

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

### $(2)$ iff $(3)$

Suppose $PQ = Q$.

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

That is, $QP = PQ = Q$.

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

$\Box$

### $(2), (3)$ imply $(1)$

Suppose $PQ = QP = Q$. Then as $P, Q$ are projections:

- $\left({P - Q}\right)^2 = P^2 + Q^2 - PQ - QP = P - Q$

That is, $P - Q$ is idempotent.

From Adjoining is Linear, $\left({P - Q}\right)^* = P^* - Q^* = P - Q$.

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:

- $\left\langle{Ph, h}\right\rangle_H - \left\langle{Qh, h}\right\rangle_H \ge 0$

Applying this statement on $P, Q$ also, one obtains that $Ph = 0 \implies Qh = 0$; that is, $\operatorname{ker} P \subseteq \operatorname{ker} Q$.

Orthocomplement Reverses Subset and that $P, Q$ are projections combine to state $\operatorname{ran} Q \subseteq \operatorname{ran} P$.

So for every $h \in H$, there is a $p \in H$ with $Qh = Pp$; it follows that $PQh = PPp = Pp = Qh$.

Hence $PQ = Q$.

$\blacksquare$

## Also see

## Sources

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