Projections

Definition 2.12: An operator $P \in \mathcal{B}(X)$ is called a projection if it is idempotent, i.e. $P^2=P$.

Definition 2.13: A self-adjoint projection $P \in \mathcal{B}(\mathcal{H})$ i.e. $P=P^*$ is called an orthogonal projection.

Lemma 2.14: Let $P \in \mathcal{B}(X)$ be a projection. Then $Q \coloneqq I-P$ is a projection, $PQ=QP=0$ and:

In particular, $\text{Ran}(P)$ is closed.

Proof:

First we see that $Q$ is a projection. Indeed:

Then we have that:

and from a similar computation $QP=0$ follows. Since $QP=0$, we have found that $\text{Ran}(P) \subset \ker(Q)$. On the other hand for any $x \in \ker(Q)$ we have that $x-Px=0$, i.e. $x=Px$, so that $x \in \text{Ran}(P)$. Thus $\ker(Q) \subset \text{Ran}(P)$, and $\text{Ran}(P) = \ker(Q)$. The equality $\ker(P) = \text{Ran}(P)$ follows from reversing the roles of $P$ and $Q$ and observing that $P=I-Q$. $\blacksquare$

Corollary 2.15: There is a one-to-one correspondence between orthogonal projections and closed subspaces, associating subspaces $M \subset \mathcal{H}$ with $P_M$ the projection on $M$ and a projection $P$ with the subspace $M_P = \text{Ran}(P)$. We have for every subspace $M \subset \mathcal{H}$:

Proof: Lemma 2.14 is one side of the implication, associating a subspace with the closed subspace. For the other side, the projection theorem tells us that for closed subspaces $\mathcal{H} = M \oplus M^{\perp}$, so that any $x \in \mathcal{H}$ can be written uniquely as $x=y+z$, $y \in M$ and $z \in M^{\perp}$. The map $P_M : x \mapsto y$ is an orthogonal projection with $\text{Ran}(P_M) = M$ (verify orthogonality), and we can apply Lemma 2.14 to it. $\blacksquare$