Geometry and algebra of projectors

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 $\text{Img}(P)\perp N(P).$