# 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

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## 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.

Also from Adjoint of Composition of Linear Transformations is Composition of Adjoints:

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

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 $P Q = Q P$.

$\blacksquare$

## Also see

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*... (previous) ... (next) $\text {II}.3$ Exercises $4, 7$