9
Nov 18

## 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 $Q.$