# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 13

    Projections
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 which 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.
$\Box$
    Spaces of homomorphisms

## Definition

Let $K$ be a field, and let $V$ and $W$ be $K$-vector spaces. Then
    $\operatorname {Hom} _{K}{\left(V,W\right)}={\left\{f:V\rightarrow W\mid f{\text{ linear mapping}}\right\}}\,$
is called the space of homomorphisms. It is endowed with the addition defined by
    $(f+g)(v):=f(v)+g(v)\,,$
and the scalar multiplication defined by
    $(\lambda f)(v):=\lambda \cdot f(v)\,.$
Due to exercise, this is indeed a $K$-vector space.

## Example

Let $W$ be a $K$-vector space over the field $K$. Then the mapping
    $\operatorname {Hom} _{K}{\left(K,W\right)}\longrightarrow W,\varphi \longmapsto \varphi (1),$
is an isomorphism of vector spaces, see exercise.
The space of homomorphisms $\operatorname {Hom} _{K}{\left(V,K\right)}$ plays an important role. It is called the dual space of $V$. In the next lectures, we will discuss it in more details.

## Lemma

Let $K$ be a field, and let $V$ and $W$ be
$K$
* A linear mapping from another vector space $U$ induces a linear mapping
    $\varphi \colon U\longrightarrow V$
    $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Hom} _{K}{\left(U,W\right)},f\longmapsto f\circ \varphi .$
* A linear mapping to another vector space $T$ induces a linear mapping
    $\psi \colon W\longrightarrow T$
    $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Hom} _{K}{\left(V,T\right)},f\longmapsto \psi \circ f.$

### Proof

$\Box$

## Lemma

Let $V$ be a $K$-vector space together with a direct sum decomposition
    $V=U_{1}\oplus U_{2}\,.$
Let $W$ be another $K$-vector space and let
    $\varphi _{1}\colon U_{1}\longrightarrow W$
and
    $\varphi _{2}\colon U_{2}\longrightarrow W$
denote linear mappings. Then we get, by setting
    $\varphi (v):=\varphi _{1}(v_{1})+\varphi _{2}(v_{2})\,,$
where $v=v_{1}+v_{2}$ is the direct decomposition, a linear mapping
    $\varphi \colon V\longrightarrow W.$

### Proof

The mapping is well-defined, since the representation $v=v_{1}+v_{2}$ with $v_{1}\in U_{1}$ and $v_{2}\in U_{2}$ is unique. The linearity follows from
    ${\begin{aligned}\varphi (av+bv')&=\varphi _{1}{\left((av+bv')_{1}\right)}+\varphi _{2}{\left((av+bv')_{2}\right)}\\&=\varphi _{1}{\left(av_{1}+bv'_{1}\right)}+\varphi _{2}{\left(av_{2}+bv'_{2}\right)}\\&=a\varphi _{1}(v_{1})+b\varphi _{1}(v'_{1})+a\varphi _{2}(v_{2})+b\varphi _{2}(v'_{2})\\&=a\varphi _{1}(v_{1})+a\varphi _{2}(v_{2})+b\varphi _{1}(v'_{1})+b\varphi _{2}(v'_{2})\\&=a\varphi (v)+b\varphi (v')\end{aligned}}$
$\Box$

## Lemma

Let $K$ be a field, and let $V$ and $W$ be $K$-vector spaces. Let
    $V=V_{1}\oplus \cdots \oplus V_{n}\,$
and
    $W=W_{1}\oplus \cdots \oplus W_{m}\,$
de direct sum decompositions and let
    $p_{j}\colon W\longrightarrow W_{j}$
denote the canonical projections. Then the mapping
    $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \prod _{1\leq i\leq n,\,1\leq j\leq m}\operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)},f\longmapsto p_{j}\circ {\left(f{|}_{V_{i}}\right)},$
is an isomorphism. If we consider $\operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)}$ as linear subspaces of $\operatorname {Hom} _{K}{\left(V,W\right)}$, then we have the direct sum decomposition
    $\operatorname {Hom} _{K}{\left(V,W\right)}\cong \bigoplus _{1\leq i\leq n,\,1\leq j\leq m}\operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)}\,.$

### Proof

It follows directly from fact that the given mapping is linear. In order to prove injectivity, let $f\in \operatorname {Hom} _{K}{\left(V,W\right)}$ with $f\neq 0$ be given. Then there exists some $v\in V$ such that
    $f(v)\neq 0\,.$
Let $v=v_{1}+\cdots +v_{n}$ with $v_{i}\in V_{i}$. Then also $f(v_{i})\neq 0$ for some $i$. Therefore, $(f(v_{i}))_{j}\neq 0$ for some $j$. Hence
    $f_{ij}=p_{j}\circ {\left(f{|}_{V_{i}}\right)}\neq 0\,.$
In order to prove surjectivity, let a family of homomorphisms $f_{ij}\in \operatorname {Hom} _{K}{\left(V_{i},W_{j}\right)},\,1\leq i\leq n,\,1\leq j\leq m$, be given, which we consider as mappings to $W$. Then the
    $f_{i}=\sum _{j=1}^{m}f_{ij}\,$
are linear mappings from $V_{i}$ to $W$. This yields via fact a linear mapping $\sum _{i=1}^{n}f_{i}$ from $V$ to $W$, which restricts to the given mappings.
$\Box$

## Theorem

Let $K$ denote a field and let $V$ and $W$ denote finite-dimensional $K$-vector spaces. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ be a basis of $V$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{m}$ be a basis of $W$. Then the assignment
    $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Mat} _{m\times n}(K),f\longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(f),$
is an isomorphism
$K$

### Proof

The bijectivity was shown in fact. The additivity follows from
    ${\left(M_{\mathfrak {w}}^{\mathfrak {v}}(f+g)\right)}_{ij}=((f+g)(v_{j}))_{i}=(f(v_{j})+g(v_{j}))_{i}=f(v_{j})_{i}+g(v_{j})_{i}={\left(M_{\mathfrak {w}}^{\mathfrak {v}}(f)\right)}_{ij}+{\left(M_{\mathfrak {w}}^{\mathfrak {v}}(g)\right)}_{ij}\,,$
where the index $i$ denotes the $i$-th component with respect to the basis ${\mathfrak {w}}$.
$\Box$
This means that we can consider the direct sum decomposition for the one-dimensional linear subspaces $Kv_{i}$ bzw. $Kw_{j}$ corresponding to bases and apply fact.

## Corollary

Let $K$ denote a field and let $V$ and $W$ denote finite-dimensional $K$-vector spaces with dimensions $n$ bzw. $m$. Then
    $\dim _{\left(\operatorname {Hom} _{K}{\left(V,W\right)}\right)}=nm\,.$

### Proof

This follows immediately from fact.
$\Box$
    Linear subspaces of homomorphism spaces

## Lemma

Let $V$ and $W$ be vector spaces over a field $K$. Then the following subsets are linear subspaces
$\operatorname {Hom} _{K}{\left(V,W\right)}$
* For a linear subspace $U\subseteq V$, is a linear subspace of $\operatorname {Hom} _{K}{\left(V,W\right)}$. If $V$ and $W$ are finite-dimensional, then
    $S={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi {|}_{U}=0\right\}}\,$
    $\dim _{\left(S\right)}={\left(\dim _{\left(V\right)}-\dim _{\left(U\right)}\right)}\dim _{\left(W\right)}\,.$
* For a linear subspace $U\subseteq W$, is a linear subspace of $\operatorname {Hom} _{K}{\left(V,W\right)}$, which is isomorphic to $\operatorname {Hom} _{K}{\left(V,U\right)}$. If $V$ and $W$ are finite-dimensional, then
    $T={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \operatorname {Im} \varphi \subseteq U\right\}}\,$
    $\dim _{\left(T\right)}=\dim _{\left(V\right)}\cdot \dim _{\left(U\right)}\,.$
* For linear subspaces $V_{1}\subseteq V$ and $W_{1}\subseteq W$, is a linear subspace of $\operatorname {Hom} _{K}{\left(V,W\right)}$. If $V$ and $W$ finite-dimensional, then
    $H={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{1})\subseteq W_{1}\right\}}\,$
    $\dim _{\left(H\right)}=\dim _{\left(V_{1}\right)}\dim _{\left(W_{1}\right)}+{\left(\dim _{\left(V\right)}-\dim _{\left(V_{1}\right)}\right)}\dim _{\left(W\right)}\,.$
* For linear subspaces $V_{1},\ldots ,V_{n}\subseteq V$ and $W_{1},\ldots ,W_{n}\subseteq W$, is a linear subspace of $\operatorname {Hom} _{K}{\left(V,W\right)}$.
    $L={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{1})\subseteq W_{1}{\text{ and }}\varphi (V_{2})\subseteq W_{2}{\text{ and }}\ldots {\text{ and }}\varphi (V_{n})\subseteq W_{n}\right\}}\,$

### Proof

(1). It is clear that we have a linear subspace. In order to prove the statement about the dimension, let $U'$ be an direct complement of $U$ in $V$, so that
    $V=U\oplus U'\,.$
Let $u_{1},\ldots ,u_{r}$ be a basis of $U'$. Every linear mapping from $S$ maps $U$ to $0$, and on $U'$ (or on a basis thereof) we have free choice. Therefore
    $S\cong \operatorname {Hom} _{K}{\left(U',W\right)}\,$
and the statement follows from fact.
(2). It is clear that we have a linear subspace. The natural mapping
    $\operatorname {Hom} _{K}{\left(V,U\right)}\longrightarrow \operatorname {Hom} _{K}{\left(V,W\right)}$
of fact is injective in this case. Therefore,
    $T\cong \operatorname {Hom} _{K}{\left(V,U\right)}\,.$
(3). It is clear that we have a linear subspace. In the finite-dimensional case, let
    $V=V_{1}\oplus V_{2}\,$
be a direct sum decomposition. Due to fact, we have
    $\operatorname {Hom} _{K}{\left(V,W\right)}=\operatorname {Hom} _{K}{\left(V_{1},W\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,$
and
    $H=\operatorname {Hom} _{K}{\left(V_{1},W_{1}\right)}\oplus \operatorname {Hom} _{K}{\left(V_{2},W\right)}\,.$
Therefore, the dimension equals
    $\dim _{\left(V_{1}\right)}\cdot \dim _{\left(W_{1}\right)}+\dim _{\left(V_{2}\right)}\cdot \dim _{\left(W\right)}=\dim _{\left(V_{1}\right)}\dim _{\left(W_{1}\right)}+{\left(\dim _{\left(V\right)}-\dim _{\left(V_{1}\right)}\right)}\dim _{\left(W\right)}\,.$
(4). Setting
    $L_{i}={\left\{\varphi \in \operatorname {Hom} _{K}{\left(V,W\right)}\mid \varphi (V_{i})\subseteq W_{i}\right\}}\,,$
we have $L=\bigcap _{i=1}^{k}L_{i}$. Hence, (4) follows from (3).
$\Box$

## Remark

Let $K$ denote a field and let $V$ and $W$ denote finite-dimensional $K$-vector spaces. We consider the natural mapping
    $\Psi \colon V\times \operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow W,(v,\varphi )\longmapsto \varphi (v),$
where we have the product space on the left. This mapping is not linear in general. We have, on one hand,
    $\varphi (v_{1}+v_{2})=\varphi (v_{1})+\varphi (v_{2})\,$
and, on the other hand,
    ${\left(\varphi _{1}+\varphi _{2}\right)}(v)=\varphi _{1}(v)+\varphi _{2}(v)\,,$
so, if we fix one component, we have additivity (and also compatibility with the scalar multiplication) in the other component. In the product space, we have
    $(v_{1},\varphi _{1})+(v_{2},\varphi _{2})=(v_{1}+v_{2},\varphi _{1}+\varphi _{2})\,,$
and, therefore,
    ${\begin{aligned}\Psi ((v_{1},\varphi _{1})+(v_{2},\varphi _{2}))&=\Psi ((v_{1}+v_{2},\varphi _{1}+\varphi _{2}))\\&={\left(\varphi _{1}+\varphi _{2}\right)}{\left(v_{1}+v_{2}\right)}\\&=\varphi _{1}{\left(v_{1}+v_{2}\right)}+\varphi _{2}{\left(v_{1}+v_{2}\right)}\\&=\varphi _{1}(v_{1})+\varphi _{1}(v_{2})+\varphi _{2}(v_{1})+\varphi _{2}(v_{2})\\&\neq \varphi _{1}(v_{1})+\varphi _{2}(v_{2})\\&=\Psi ((v_{1},\varphi _{1}))+\Psi ((v_{2},\varphi _{2}))\end{aligned}}$
(only in exceptional cases we have $\varphi _{1}(v_{2})+\varphi _{2}(v_{1})=0$).
    Footnotes
* ↑ These projections are even orthogonal projections. This term requires that an inner product is defined, a concept we will discuss in the second term. Here, we only deal with linear projections, which depend, different from the orthogonal case, from the direct complement chosen.