# projection

###### Proposition 1

If $P:V\rightarrow V$ is a projection, then its image and the kernel are complementary subspaces, namely

 $V=\ker P\oplus\mathop{\mathrm{img}}\nolimits P.$ (2)

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

 $u=v-Pv.$

The projection condition (1) 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

 $V=V_{1}\oplus V_{2}$

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 $\mathbb{R}$ or $\mathbb{C}$ and that $V$ is equipped with a positive-definite inner product. In this setting we call an endomorphism   $P:V\rightarrow V$ an orthogonal projection if it is self-dual

 $P^{\displaystyle\star}=P,$
###### Proposition 2

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

Proof. Let $u\in\ker P$ and $v\in\mathop{\mathrm{img}}\nolimits P$ 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 $u\in\mathop{\mathrm{img}}\nolimits P$.

Title projection Projection 2013-03-22 12:52:13 2013-03-22 12:52:13 rmilson (146) rmilson (146) 8 rmilson (146) Definition msc 15A21 msc 15A57 orthogonal projection