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Operators on Infinite Dimensional Vector Spaces 6

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Projections

Definition 2.12: An operator PB(X)P \in \mathcal{B}(X) is called a projection if it is idempotent, i.e. P2=PP^2=P.

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

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

Ran(P)=ker(Q), and Ran(Q)=ker(P).\text{Ran}(P) = \ker(Q), \ \text{and} \ \text{Ran}(Q) = \ker(P).

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

Proof:

First we see that QQ is a projection. Indeed:

Q2=(IP)2=I2P+P2=IP=Q.Q^2 = (I-P)^2 = I-2P + P^2 = I-P=Q.

Then we have that:

PQ=P(IP)=PP2=0,PQ = P(I-P) = P-P^2=0,

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

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

H=MM=Ran(PM)Ran(IPM).\mathcal{H} = M \oplus M^{\perp} = \text{Ran}(P_M) \oplus \text{Ran}(I-P_M).

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 H=MM\mathcal{H} = M \oplus M^{\perp}, so that any xHx \in \mathcal{H} can be written uniquely as x=y+zx=y+z, yMy \in M and zMz \in M^{\perp}. The map PM:xyP_M : x \mapsto y is an orthogonal projection with Ran(PM)=M\text{Ran}(P_M) = M (verify orthogonality), and we can apply Lemma 2.14 to it. \blacksquare