projection
Projection
2002-07-26 07:46:53
2004-02-25 11:15:48
Definition
orthogonal projection
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A linear transformation $P:V\rightarrow V$ of a vector space $V$ is called a
\emph{projection} if it acts like the identity on its image. This
condition can be more succinctly expressed by the equation
\begin{equation}
\label{eq:proj}
P^2 = P.
\end{equation}
\begin{proposition}
If $P:V\rightarrow V$ is a projection, then
its image and the kernel are complementary subspaces, namely
\begin{equation}
\label{eq:comp}
V = \ker P \oplus \img P.
\end{equation}
\end{proposition}
\emph{Proof. }
Suppose that $P$ is a projection. Let $v\in V$ be given, and set
$$u=v-Pv.$$
The projection condition \eqref{eq:proj} then implies
that $u\in \ker P$, 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 $v\in V$ 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 $P:V\rightarrow V$ defined by
$$
Pv =\begin{cases}
v & v\in V_1 \\
0 & v\in V_2
\end{cases}$$
Specializing somewhat, suppose that the ground field is $\reals$ or
$\cnums$ and that $V$ is equipped with a positive-definite inner
product. In this setting we call an endomorphism
$P:V\rightarrow V$ an \emph{orthogonal projection} if it is self-dual
$$P\adj = P,$$
in addition to satisfying the projection condition \eqref{eq:proj}.
\begin{proposition}
The kernel and image of an orthogonal projection are orthogonal subspaces.
\end{proposition}
\emph{Proof. }
Let $u\in\ker P$ and $v\in \img P$ be given. Since $P$
is self-dual we have
$$0 = \langle Pu,v\rangle = \langle u,Pv\rangle = \langle
u,v\rangle.$$
QED
Thus we see that a orthogonal projection $P$ projects a $v \in V$ onto
$Pv$ in an orthogonal fashion, i.e.
$$\langle v-Pv,u\rangle = 0$$
for all $u\in \img P$.