projection
A linear transformation P:V→V of a vector space V is called a
projection if it acts like the identity
on its image. This
condition can be more succinctly expressed by the equation
P2=P. | (1) |
Proposition 1
If P:V→V is a projection, then its image and the kernel are complementary subspaces, namely
V=kerP⊕imgP. | (2) |
Proof. Suppose that P is a projection. Let v∈V be given, and set
u=v-Pv. |
The projection condition (1) then implies that u∈kerP, and we can write v as the sum of an image and kernel vectors:
v=u+Pv. |
This decomposition is unique, because the
intersection 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, Pv=v and Pv=0, and hence v=0. QED
Conversely, every direct sum decomposition
V=V1⊕V2 |
corresponds to a projection P:V→V defined by
Pv={vv∈V10v∈V2 |
Specializing somewhat, suppose that the ground field is ℝ or
ℂ and that V is equipped with a positive-definite inner
product. In this setting we call an endomorphism
P:V→V an orthogonal projection if it is self-dual
P⋆=P, |
Proposition 2
The kernel and image of an orthogonal projection are orthogonal subspaces.
Proof. Let u∈kerP and v∈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∈V onto
Pv in an orthogonal fashion, i.e.
⟨v-Pv,u⟩=0 |
for all u∈imgP.
Title | projection |
---|---|
Canonical name | Projection |
Date of creation | 2013-03-22 12:52:13 |
Last modified on | 2013-03-22 12:52:13 |
Owner | rmilson (146) |
Last modified by | rmilson (146) |
Numerical id | 8 |
Author | rmilson (146) |
Entry type | Definition |
Classification | msc 15A21 |
Classification | msc 15A57 |
Defines | orthogonal projection |