Orthogonal Projections: Definition + Characterization

Question

Hey there I'm hanging at so many little steps when proving:
$P\text{ orthogonal projection}\iff X\text{ orthogonal decomposable}$

My problem is there is simply missing a precise definition of what an orthogonal projection should be. For example, when introducing orthogonal projections first time they pop up from the orthogonal decomposition of complete(!) subspaces. However, that I noticed is not really necessary to have a decomposition into orthogonal subspaces. That is just one out of many subtleties (linearity, continuity, etc.)...

So my question:
$P\text{ orthogonal projection}:\iff\text{???}$

Answer 1

Characterizartion:

There's a one to one correspondence between linear decompositions and linear projections: $$\left(X=U\oplus V\right)\leftrightarrow\left(P:X\to X:P^2=P\text{ linear}\right)$$

and a one to one correspondence between orthogonal decompositions and orthogonal projections: $$\left(X=U\underline{\oplus}V\right)\leftrightarrow\left(P:X\to X:P^2=P=P^*\text{ linear}\right)$$

Remark:

An orthogonal projection is necessarily continuous since: $$\lVert x\rVert^2=\lVert Px\rVert^2+\lVert x-Px\rVert^2$$

Warning:

There's no one to one correspondence between nonlinear decompositions and nonlinear projections: $$\pi:\mathcal{R}\to\mathcal{R}:\pi(x\neq 0)=\frac{x}{\lvert x\rvert},\pi(x=0)=0$$

Answer 2

Let $V$ be a vectorspace. Suppose we can write $V = G\oplus H$, then we can define a function $P:V\rightarrow G$ by sending $v = g+h\mapsto g$. The function $P$ is called the projection onto $G$. If $G = H^\bot$, so the orthogonal complement of $H$, then we get $V=H\oplus H^\bot$, an orthogonal decomposition of $V$, in which case $P$ is called an orthogonal projection.

Answer 3

If $H$ is a Hilbert space, $P:H \rightarrow H$ is an orthogonal projection if there exists a closed subspace $M$ of $H$ such that $Pv = v$ for $v \in M$ and $Pw = 0$ for $w \in M^{\perp}$ ($M^{\perp}$ is the orthogonal complement of $M$).

Answer 4

The orthogonal projection $P_Y$ of some space $X$ (finite-dimensional) onto some subspace $Y$ is defined in such a way that $P_ux=y$ where $x= y + y'$ for $y \in Y$ and $y'\in Y^{\perp}$. Do you know the reason why such a decomposition is possible?

Warning: I am talking about basic linear algebra here, so ignore if I am out of my depth.

Answer 5

Hm, why not say that a space with scalar product $V$ is decomposed into $V=R\oplus S$. We define a linear operator $P:V\to V$ such that $P(r+s) = r$. Then we say that it is orthogonal if $$\forall v\in V \quad (I-P)v\bot Pv.$$

Will this definition work for you?

Answer 6

Here's a nice definition:
$T\text{ orthogonal}:\iff\mathcal{R}(T)\bot\mathcal{N}(T)$

The nice thing on this definition is that it does neither privilege the range nor the kernel.
Moreover it applies to a more general class of operators as to nonlinear, unbounded, etc. in pre hilbert spaces.
For the special class of linear projections this reduces to the well known orthogonal decomposition.