# Product 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
$(2): \quad P Q = Q P$
$(3): \quad P + Q - P Q$ is a projection

## Proof

The proof proceeds by first showing that $(1)$ is equivalent to $(2)$.

Then, these are combined and shown equivalent to $(3)$.

### $(1)$ implies $(2)$

Let $P Q$ be a projection.

Then by Characterization of Projections, statement $(4)$, one has:

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

where the penultimate equality follows from Adjoint of Composition of Linear Transformations is Composition of Adjoints.

$\Box$

### $(2)$ implies $(1)$

Let $P Q = Q P$.

Then:

$\paren {P Q}^2 = P Q P Q = P^2 Q^2 = P Q$

as $P, Q$ are projections.

Hence $P Q$ is an idempotent.

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

Hence, by Characterization of Projections, statement $(4)$, $P Q$ is a projection.

$\Box$

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

The above establishes that assuming either of $(1)$ and $(2)$ yields both to hold.

So assuming $(1)$, $P, Q$ and $P Q$ are all projections, and $P Q = Q P$.

Now compute:

 $\ds \paren {P + Q - P Q}^2$ $=$ $\ds P^2 + Q^2 + \paren {P Q}^2 + P Q + Q P - P P Q - P Q P - Q P Q - P Q Q$ $\ds$ $=$ $\ds P + Q - P Q + P Q + P Q - P Q - P Q - P Q - P Q$ $P, Q, P Q$ projections, $P Q = Q P$ $\ds$ $=$ $\ds P + Q - P Q$

It follows that $P + Q - P Q$ is an idempotent.

From:

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

Now applying Characterization of Projections, statement $(4)$, conclude that $P + Q - P Q$ is a projection.

$\Box$

### $(3)$ implies $(2)$

Let $P + Q - P Q$ be a projection.

Then by Characterization of Projections, statement $(4)$, compute:

 $\ds P + Q - P Q$ $=$ $\ds \paren {P + Q - P Q}^*$ $\ds$ $=$ $\ds P^* + Q^* - Q^* P^*$ Adjoining is Linear, Adjoint of Composition of Linear Transformations is Composition of Adjoints $\ds$ $=$ $\ds P + Q - Q P$

Hence necessarily $P Q = Q P$.

$\blacksquare$