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 ${}\varphi {|}_{U}=\operatorname {Id} _{U}$

holds.

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

${}\varphi ^{2}=\varphi \,$

holds.

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

${}V=\operatorname {kern} \varphi \oplus \operatorname {Im} \varphi \,,$

and ${}\varphi$ is the projection onto ${}\operatorname {Im} \varphi$.

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.

$\Box$