projection

A linear transformation $P:V\to V$ of a vector space $V$ is called aprojection if it acts like the identity on its image. Thiscondition can be more succinctly expressed by the equation

$$P^2=P.$$(1)

Proposition 1

If $P\mathrm{:}V\mathrm{\to }V$ is a projection, thenits image and the kernel are complementary subspaces, namely

$$V=\ker P\oplus imgP.$$(2)

Proof. Suppose that $P$ is a projection. Let $v\in V$ be given, and set

$$u=v-Pv.$$

The projection condition (1) then impliesthat $u\in \ker P$, and we can write $v$ as the sum of an image andkernel vectors:

$$v=u+Pv.$$

This decomposition is unique, because theintersection of the image and the kernel is the trivial subspace.Indeed, suppose that $v\in V$ is in both the image and the kernel of $P$.Then, $Pv=v$ and $Pv=0$, and hence $v=0$. QED

Conversely, every direct sum decomposition

$$V=V_1\oplus V_2$$

corresponds to a projection $P:V\to V$ defined by

Specializing somewhat, suppose that the ground field is $ℝ$ or$ℂ$ and that $V$ is equipped with a positive-definite innerproduct. In this setting we call an endomorphism$P:V\to V$ an orthogonal projection if it is self-dual

$$P^{\star }=P,$$

in addition to satisfying the projection condition (1).

Proposition 2

The kernel and image of an orthogonal projection are orthogonal subspaces.

Proof. Let $u\in \ker P$ and $v\in imgP$ be given. Since $P$is self-dual we have

$$0=⟨Pu,v⟩=⟨u,Pv⟩=⟨u,v⟩.$$

QED

Thus we see that a orthogonal projection $P$ projects a $v\in V$ onto$Pv$ in an orthogonal fashion, i.e.

$$⟨v-Pv,u⟩=0$$

for all $u\in imgP$.