# Vector space/Projection/Introduction/Section

For a direct sum decomposition $V=U_{1}\oplus U_{2}$ of a $K$-vector space $V$, the linear mapping
    $p_{1}\colon V\longrightarrow U_{1},v_{1}+v_{2}\longmapsto v_{1},$
is called the first projection (or projection onto $U_{1}$ with respect to the given decomposition or projection onto $U_{1}$ along $U_{2}$) and, accordingly,
    $p_{2}\colon V\longrightarrow U_{2},v_{1}+v_{2}\longmapsto v_{2},$
the second projection of this decomposition. Since $U_{1}$ and $U_{2}$ are linear subspaces of $V$, it is reasonable to call the composed mapping
    $V{\stackrel {p_{1}}{\longrightarrow }}U_{1}\longrightarrow V$
a projection as well. Then we have a projection in the sense of the following definition.

## Definition

Let $K$ be a field, $V$ a $K$-vector space and $U\subseteq V$ a linear subspace. A linear mapping
    $\varphi \colon V\longrightarrow V$
is called a projection of $V$ onto $U$, if $U=\operatorname {Im} \varphi$ and $\varphi {|}_{U}=\operatorname {Id} _{U}$

## Example

Let $V$ be a finite-dimensional $K$-vector space and $v_{i}$, $i\in I$, a basis of $V$. For a subset $J\subseteq I$, set
    $V_{J}=\langle v_{i},\,i\in J\rangle \,,$
the linear subspace corresponding to $J$. Moreover, let
    $p_{J}\colon V\longrightarrow V,\sum _{i\in I}a_{i}v_{i}\longmapsto \sum _{i\in J}a_{i}v_{i},$
the corresponding projection. The image of this projection is $V_{J}$. On $V_{J}$, this mapping is the identity, one can also consider this mapping as
    $p_{J}\colon V\longrightarrow V_{J}.$
The kernel of this mapping is
    $\operatorname {kern} p_{J}=\langle v_{i},\,i\notin J\rangle \,.$

## Example

For $\mathbb {R} ^{3}$, equipped with the standard basis, we consider subsets $J\subset \{1,2,3\}$ with two elements and the corresponding projections (in the sense of example) onto the coordinate planes. The projections are
    $p_{\{1,2\}}\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3},(a,b,c)\longmapsto (a,b,0),$
    $p_{\{1,3\}}\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3},(a,b,c)\longmapsto (a,0,c),$
    $p_{\{2,3\}}\colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3},(a,b,c)\longmapsto (0,b,c).$
The images of some object in $\mathbb {R} ^{3}$ under these projections carry names like ground view etc.
For a subset $\{j\}\subseteq \{1,2,3\}$ with one element, the projections map to an axis.
A more abstract definition is the following that does not refer to a linear subspace.

## Definition

Let $K$ be a field and $V$ a $K$-vector space. A linear mapping
    $\varphi \colon V\longrightarrow V$
is called a projection, if
    $\varphi ^{2}=\varphi \,$
The identity and the zero mapping are projections.

## Lemma

Let $K$ be a field and $V$ a $K$-vector space. For a direct sum decomposition
    $V=U\oplus U'\,,$
the projection onto $U$ is a projection in the sense of Definition. Such a projection
    $\varphi \colon V\longrightarrow V$
gives a decomposition
    $V=\operatorname {kern} \varphi \oplus \operatorname {Im} \varphi \,,$
$\varphi$
$\operatorname {Im} \varphi$

### Proof

Let $\pi _{U}$ be the projection onto $U$. Write $v=u+u'$ with $u\in U,\,u'\in U'$. Then we have
    $\pi _{U}(\pi _{U}(u+u'))=\pi _{U}(u)=u=\pi _{U}(u+u')\,,$
and hence
    $\pi _{U}^{2}=\pi _{U}\,.$
Suppose now that
    $\varphi \colon V\longrightarrow V$
is an endomorphism with
    $\varphi ^{2}=\varphi \,.$
Let $x\in \operatorname {kern} \varphi \cap \operatorname {Im} \varphi$. Then there exists some $y\in V$ such that
    $\varphi (y)=x\,.$
Then
    $x=\varphi (y)=\varphi (\varphi (y))=\varphi (x)=0\,.$
This means that the intersection of these linear subspaces is the zero space. For an arbitrary $v\in V$, we write
    $v=\varphi (v)+(v-\varphi (v))\,.$
Here, the first summand belongs to the image and, because of
    $\varphi (v-\varphi (v))=\varphi (v)-\varphi (\varphi (v))=\varphi (v)-\varphi (v)=0\,,$
the second summand belongs to the kernel. Therefore, we have a direct sum decomposition.
$\Box$