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Nov 18

Geometry and algebra of projectors




Geometry and algebra of projectors

Projectors are geometrically so simple that they should have been discussed somewhere in the beginning of this course. I am giving them now because the applications are more advanced.

Motivating example

Let L be the x-axis and L^\perp the y-axis on the plane. Let P be the projector onto L along L^\perp and let Q be the projector onto L^\perp along L. This geometry translates into the following definitions:

L=\{(x,0):x\in R\}, L^\perp=\{(0,y):y\in R\}, P(x,y)=(x,0), Q(x,y)=(0,y).

The theory is modeled on the following observations.

a) P leaves the elements of L unchanged and sends to zero all elements of L^\perp.

b) L is the image of P and L^\perp is the null space of P.

c) Any element of the image of P is orthogonal to any element of the image of Q.

d) Any x can be represented as x=(x_1,0)+(0,x_2)=Px+Qx. It follows that I=P+Q.

For more simple examples, see my post on conditional expectations.

Formal approach

Definition 1. A square matrix P is called a projector if it satisfies two conditions: 1) P^2=P (P is idempotent; for some reason, students remember this term better than others) and 2) P^T=P (P is symmetric).

Exercise 1. Denote L_P=\{x:Px=x\} the set of points x that are left unchanged by P. Then L_P is the image of P (and therefore a subspace).

Proof. Indeed, the image of P consists of points y=Px. For any such y, we have Py=P^2x=Px=y, so y belongs to L_P. Conversely, any element of L_P is seen to belong to the image of P.

Exercise 2. a) The null space and image of P are orthogonal. b) We have an orthogonal decomposition R^n=N(P)\oplus \text{Img}(P).

Proof. a) If x\in \text{Img}(P) and y\in N(P), then Py=0 and by Exercise 1 Px=x. Therefore x\cdot y=(Px)\cdot y=x\cdot (Py)=0. This shows that \text{Img}(P)\perp N(P).

b) For any x write x=Px+(I-P)x. Here Px\in \text{Img}(P) and (I-P)x\in N(P) because P(I-P)x=(P-P^2)x=0.

Exercise 3. a) Along with P, the matrix Q=I-P is also a projector. b) \text{Img}(Q)=N(P) and N(Q)=\text{Img}(P).

Proof. a) Q is idempotent: Q^2=(I-P)^2=I-2P+P^2=I-P=Q. b) Q is symmetric: Q^T=I^T-P^T=Q.

b) By Exercise 2

\text{Img}(Q)=\{x:Qx=x\}=\{x:(I-P)x=x\}=\{x:Px=0\}=N(P).

Since P=I-Q, this equation implies N(Q)=\text{Img}(P).

It follows that, as with P, the set L_Q=\{x:Qx=x\} is the image of Q and it consists of points that are not changed by Q.

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