Raising the bar

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

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

The theory is modeled on the following observations.

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

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

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

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

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

Formal approach

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

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

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

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

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

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

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

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

b) By Exercise 2

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

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

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

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