projection (original) (raw)

Proposition 1

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

The projection condition (1) then implies that u∈ker⁡P, and we can write v as the sum of an image and kernel vectors:

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

Proposition 2

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

Proof. Let u∈ker⁡P and v∈imgP be given. Since Pis self-dual we have

QED

Thus we see that a orthogonal projection P projects a v∈V ontoP⁢v in an orthogonal fashion, i.e.

for all u∈imgP.