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 (2nd ed.) ... (previous) ... (next) $\text {II}.3$ Exercises $4, 7$