# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 10

    Linear mappings
We are interested in mappings between two vector spaces which respect the structures, which are compatible with addition and with scalar multiplication.

## Definition

Let $K$ be a field, and let $V$ and $W$ be $K$-vector spaces. A mapping
    $\varphi \colon V\longrightarrow W$
is called a linear mapping, if the following two properties are fulfilled.
* $\varphi (u+v)=\varphi (u)+\varphi (v)$ for all $u,v\in V$.
* $\varphi (sv)=s\varphi (v)$ for all $s\in K$ and $v\in V$.
Here, the first property is called additivity and the second property is called compatibility with scaling. When we want to stress the base field, then we say $K$-linearity. The identity $\operatorname {Id} _{V}\colon V\rightarrow V$, the null mapping $V\rightarrow 0$ and the inclusion $U\subseteq V$ of a linear subspace are the simplest examples for linear mappings. For a linear mapping, the compatibility with arbitrary linear combination holds, that is,
    $\varphi {\left(\sum _{i=1}^{n}s_{i}v_{i}\right)}=\sum _{i=1}^{n}s_{i}\varphi (v_{i})\,,$
see exercise. Instead of linear mappings, we also say homomorphism.

## Example

The easiest
linear mappings are (beside the null mapping) the linear maps from $K$ to $K$. Such a linear mapping
    $\varphi \colon K\longrightarrow K,x\longmapsto \varphi (x),$
is determined (by fact, but this is also directly clear) by $\varphi (1)$, or by the value $\varphi (t)$ for an single element $t\in K$, $t\neq 0$. In particular, $\varphi (x)=ax$, with a uniquely determined $a\in K$. In the context of physics, for $K=\mathbb {R}$, and if there is a linear relation between two measurable quantities, we talk about proportionality, and $a$ is called the proportionality factor. In school, such a linear relation occurs as "rule of three“.
Many important functionen, in particular from $\mathbb {R}$ to $\mathbb {R}$, are not linear. For example, the squaring $x\mapsto x^{2}$, the square root, the trigonometric functions, the exponential function, the logarithm is not linear. But also for such more complicated functions there are, in the framework of differential calculus, linear approximations which help to understand these functions.

## Example

Let $K$ denote a field, and let $K^{n}$ be the $n$-dimensional standard space. Then the $i$-th projection, this is the mapping
    $K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{i-1},\,x_{i},\,x_{i+1},\,\ldots ,\,x_{n}\right)\longmapsto x_{i},$
is a $K$-linear mapping. This follows immediately from componentwise addition and scalar multiplication on the standard space. The $i$-th projection is also called the $i$-th coordinate function.

## Example

In a shop, there are $n$ different products to buy, and the price of the $i$-th product (with respect to a certain unit) is $a_{i}$. A purchase is described by the $n$-tuple
    $\left(x_{1},\,x_{2},\,\ldots ,\,x_{n}\right),$
where $x_{i}$ is the amount of the $i$-th product bought. The price for the purchase is hence $\sum _{i=1}^{n}a_{i}x_{i}$. The price mapping
    $\mathbb {R} ^{n}\longrightarrow \mathbb {R} ,\left(x_{1},\,x_{2},\,\ldots ,\,x_{n}\right)\longmapsto \sum _{i=1}^{n}a_{i}x_{i},$
is linear. This means, for example, that if we do first the purchase $\left(x_{1},\,x_{2},\,\ldots ,\,x_{n}\right)$ and then, a week later, the purchase $\left(y_{1},\,y_{2},\,\ldots ,\,y_{n}\right)$, then the price of the two purchases together is the same as the price of the purchase $\left(x_{1}+y_{1},\,x_{2}+y_{2},\,\ldots ,\,x_{n}+y_{n}\right)$.

## Example

The mapping
    $K^{n}\longrightarrow K^{m},{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\longmapsto M{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}={\begin{pmatrix}\sum _{j=1}^{n}a_{1j}s_{j}\\\sum _{j=1}^{n}a_{2j}s_{j}\\\vdots \\\sum _{j=1}^{n}a_{mj}s_{j}\end{pmatrix}},$
which is given (see example) by an $m\times n$-matrix $M=(a_{ij})_{1\leq i\leq m,\,1\leq j\leq n}$, is linear.

## Definition

Let $V$ be a vector space over a field $K$. For $a\in K$, the linear mapping
    $\varphi \colon V\longrightarrow V,v\longmapsto av,$
is called homothety (or dilation)
$a$
For a homothety, the domain space and the target space are the same. The number $a$ is called scaling factor. For $a=1$, we get the identity, for $a=-1$, we get a point reflection.

## Example

Let $C^{0}(\mathbb {R} ,\mathbb {R} )$ denote the space of continuous functions from $\mathbb {R}$ to $\mathbb {R}$ and let $C^{1}(\mathbb {R} ,\mathbb {R} )$ denote the space of continuously differentiable functions. Then the mapping
    $D\colon C^{1}(\mathbb {R} ,\mathbb {R} )\longrightarrow C^{0}(\mathbb {R} ,\mathbb {R} ),f\longmapsto f',$
which assigns to a function its derivative, is linear. In analysis, we proof that
    $(af+bg)'=af'+bg'\,$
for $a,b\in \mathbb {R}$ and another function $g\in C^{1}(\mathbb {R} ,\mathbb {R} )$ holds.

## Lemma

Let $K$ denote a field, and let $U,V,W$ denote vector spaces over $K$. Suppose that
    $\varphi :U\longrightarrow V\,\,{\text{ and }}\,\,\psi :V\longrightarrow W$
are linear mappings. Then also the composition
    $\psi \circ \varphi \colon U\longrightarrow W$

### Proof

$\Box$

## Lemma

Let $K$ be a field, and let $V$ and $W$ be $K$-vector spaces. Let
    $\varphi \colon V\longrightarrow W$
be a bijective linear map. Then also the inverse mapping
    $\varphi ^{-1}\colon W\longrightarrow V$

### Proof

$\Box$
    Determination on a basis
Behind the following statement (the determination theorem) is the important principle, that in linear algebra (of finite dimensional vector spaces), the objects are determined by finitely many data.

## Theorem

Let $K$ be a field, and let $V$ and $W$ be $K$-vector spaces. Let $v_{i}$, $i\in I$, denote a basis of $V$, and let $w_{i}$, $i\in I$, denote elements in $W$. Then there exists a unique linear mapping
    $f\colon V\longrightarrow W$
with
    $f(v_{i})=w_{i}{\text{ for all }}i\in I.$

### Proof

Since we want $f(v_{i})=w_{i}$ and since a linear mapping respects all linear combinations, that is
    $f{\left(\sum _{i\in I}s_{i}v_{i}\right)}=\sum _{i\in I}s_{i}f{\left(v_{i}\right)}\,$
holds, and since every vector $v\in V$ is such a linear combination, there can at most exist one such a linear mapping. We define now a mapping
    $f\colon V\longrightarrow W$
in the following way: we write every vector $v\in V$ with the given basis as
    $v=\sum _{i\in I}s_{i}v_{i}\,$
(where $s_{i}=0$ for almost all $i\in I$) and define
    $f(v):=\sum _{i\in I}s_{i}w_{i}\,.$
Since the representation of $v$ as such a linear combination is unique, this mapping is well-defined. Linearity. For two vectors $u=\sum _{i\in I}s_{i}v_{i}$ and $v=\sum _{i\in I}t_{i}v_{i}$, we have
    ${\begin{aligned}f{\left(u+v\right)}&=f{\left({\left(\sum _{i\in I}s_{i}v_{i}\right)}+{\left(\sum _{i\in I}t_{i}v_{i}\right)}\right)}\\&=f{\left(\sum _{i\in I}{\left(s_{i}+t_{i}\right)}v_{i}\right)}\\&=\sum _{i\in I}(s_{i}+t_{i})f{\left(v_{i}\right)}\\&=\sum _{i\in I}s_{i}f{\left(v_{i}\right)}+\sum _{i\in I}t_{i}f(v_{i})\\&=f{\left(\sum _{i\in I}s_{i}v_{i}\right)}+f{\left(\sum _{i\in I}t_{i}v_{i}\right)}\\&=f(u)+f(v).\end{aligned}}$
The compatibility with scalar multiplication is shown in a similar way, see exercise.
$\Box$
In particular, a linear mapping $\varphi \colon K^{n}\rightarrow K^{m}$ is uniquely determined by $\varphi (e_{1}),\ldots ,\varphi (e_{n})$.

## Example

In many situations, a certain object (like a cube) in space $\mathbb {R} ^{3}$ shall be drawn in the plane $\mathbb {R} ^{2}$. One possibility is to work a projection. This is a linear mapping
    $\mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{2}$
which is given (with respect to the standard bases $e_{1},e_{2},e_{3}$ and $f_{1},f_{2}$) by
    $e_{1}\longmapsto f_{1},\,e_{2}\longmapsto af_{1}+bf_{2},\,e_{3}\longmapsto f_{2},$
where the coefficients $a,b$ are usually chosen in the range $[{\frac {1}{3}},{\frac {1}{2}}]$. Linearity has the effect that parallel lines are mapped to parallel lines (unless they are mapped to a point). The point $(x,y,z)$ is mapped to $(x+ay,by+z)$. The image of the object under such a linear mapping is called a projection image.
    Linear mappings and matrices
A linear mapping
    $\varphi \colon K^{n}\longrightarrow K^{m}$
is determined uniquely by the images $\varphi (e_{j})$, $j=1,\ldots ,n$, of the standard vectors, and every $\varphi (e_{j})$ is a linear combination
    $\varphi (e_{j})=\sum _{i=1}^{m}a_{ij}e_{i}\,$
and hence determined by the elements $a_{ij}$. This means all together that such a linear mapping is given by the $mn$ elements $a_{ij}$, $1\leq i\leq m$, $1\leq j\leq n$. Such a set of data can be written as a matrix. Due to fact, this observation holds for all finite-dimensional vector spaces, as long as bases are fixed on the source space and on the target space of the linear mapping.

## Definition

Let $K$ denote a field, and let $V$ be an $n$-dimensional vector space with a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$, and let $W$ be an $m$-dimensional vector space with a basis ${\mathfrak {w}}=w_{1},\ldots ,w_{m}$.
For a linear mapping
    $\varphi \colon V\longrightarrow W,$
the matrix
    $M=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )=(a_{ij})_{ij}\,,$
where $a_{ij}$ is the $i$-th coordinate of $\varphi (v_{j})$ with respect to the basis ${\mathfrak {w}}$, is called the describing matrix for $\varphi$ with respect to the bases.
For a matrix $M=(a_{ij})_{ij}\in \operatorname {Mat} _{m\times n}(K)$, the linear mapping $\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)$ determined by
    $v_{j}\longmapsto \sum _{i=1}^{m}a_{ij}w_{i}$
in the sense of fact,
$M$
For a linear mapping $\varphi \colon K^{n}\rightarrow K^{m}$, we always assume that everything is with respect to the standard bases, unless otherwise stated. For a linear mapping $\varphi \colon V\rightarrow V$ from a vector space in itself (what is called an endomorphism), one usually takes the same bases on both sides. The identity on a vector space of dimension $n$ is described by the identity matrix, with respect to every basis.
If $V=W$, then we are usually interested in the describing matrix with respect to one basis ${\mathfrak {v}}$ of $V$.

## Example

Let $V$ denote a vector space with bases ${\mathfrak {v}}$ and ${\mathfrak {w}}$. If we consider the identity
    $\operatorname {Id} \colon V\longrightarrow V$
with respect to the basis ${\mathfrak {v}}$ on the source and the basis ${\mathfrak {w}}$ on the target, we get, because of
    $\operatorname {Id} (v_{j})=v_{j}=\sum a_{ij}w_{i}\,,$
directly
    $M_{\mathfrak {w}}^{\mathfrak {v}}(\operatorname {Id} )=M_{\mathfrak {w}}^{\mathfrak {v}}\,.$
This means that the describing matrix of the identical linear mapping is the transformation matrix for the base change from ${\mathfrak {v}}$ to ${\mathfrak {w}}$.

## Lemma

Let $K$ denote a field and let $V$ denote an $n$-dimensional vector space with a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$. Let $W$ be an $m$-dimensional vector space with a basis ${\mathfrak {w}}=w_{1},\ldots ,w_{m}$, and let
    $\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V$
and
    $\Psi _{\mathfrak {w}}\colon K^{m}\longrightarrow W$
be the corresponding mappings. Let
    $\varphi \colon V\longrightarrow W$
denote a linear mapping with describing matrix $M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$. Then
    $\varphi \circ \Psi _{\mathfrak {v}}=\Psi _{\mathfrak {w}}\circ M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\,$
hold, that is, the diagram
    ${\begin{matrix}K^{n}&{\stackrel {\Psi _{\mathfrak {v}}}{\longrightarrow }}&V&\\\!\!\!\!\!M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )\downarrow &&\downarrow \varphi \!\!\!\!\!&\\K^{m}&{\stackrel {\Psi _{\mathfrak {w}}}{\longrightarrow }}&W&\!\!\!\!\!\\\end{matrix}}$
commutes. For a vector $v\in V$,
$\varphi (v)$
$v$
${\mathfrak {v}}$
$M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )$
$m$
${\mathfrak {w}}$

### Proof

$\Box$

## Theorem

Let $K$ be a field, and let $V$ be an $n$-dimensional vector space with a basis ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$, and let $W$ be an $m$-dimensional vector space with a basis ${\mathfrak {w}}=w_{1},\ldots ,w_{m}$. Then the mappings
    $\varphi \longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ){\text{ and }}M\longmapsto \varphi _{\mathfrak {w}}^{\mathfrak {v}}(M),$
defined in definition, are inverse

### Proof

We show that both compositions are the identity. We start with a matrix $M={\left(a_{ij}\right)}_{ij}$ and consider the matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)).$
Two matrices are equal, when the entries coincide for every index pair $(i,j)$. We have
    ${\begin{aligned}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)))_{ij}&=i-{\text{th coordinate of }}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M))(v_{j})\\&=i-{\text{th coordinate of }}\sum _{i=1}^{m}a_{ij}w_{i}\\&=a_{ij}.\end{aligned}}$
Now, let $\varphi$ be a linear mapping, we consider
    $\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )).$
Two linear mappings coincide, due to fact, when they have the same values on the basis $v_{1},\ldots ,v_{n}$. We have
    $(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )))(v_{j})=\sum _{i=1}^{m}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))_{ij}\,w_{i}\,.$
Due to the definition, the coefficient $(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))_{ij}$ is the $i$-th coordinate of $\varphi (v_{j})$ with respect to the basis $w_{1},\ldots ,w_{m}$. Hence, this sume equals $\varphi (v_{j})$.
$\Box$
We denote the set of all linear mappings from $V$ to $W$ by $\operatorname {Hom} _{K}{\left(V,W\right)}$. fact means that the mapping
    $\operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow \operatorname {Mat} _{m\times n}(K),\varphi \longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ),$
is bijective with the given inverse mapping. A lineare Mapping
    $\varphi \colon V\longrightarrow V$
is called an endomorphism. The set of all endomorphisms on $V$ is denoted by $\operatorname {End} _{K}{\left(V\right)}$.
    Isomorphic vector spaces

## Definition

Let $K$ denote a field and let $V$ and $W$ denote $K$-vector spaces. A bijective linear mapping
    $\varphi \colon V\longrightarrow W$
An isomorphism from $V$ to $V$ is called automorphism.

## Definition

Let $K$ denote a field. Two $K$-vector spaces $V$ and $W$ are called isomorphic, if there exists an isomorphism
$V$
$W$

## Theorem

Let $K$ denote a field and let $V$ and $W$ denote finite-dimensional $K$-vector spaces. Then $V$ and $W$ are isomorphic to each other if and only if their dimension
$n$
$K$
$K^{n}$

### Proof

$\Box$

## Remark

An isomorphism between an $n$-dimensional vector space $V$ and the standard space $K^{n}$ is essentially equivalent with the choice of a basis of $V$. For a basis
    ${\mathfrak {v}}=v_{1},\ldots ,v_{n}\,,$
we associate the linear mapping
    $\Psi _{\mathfrak {v}}\colon K^{n}\longrightarrow V,e_{i}\longmapsto v_{i},$
which maps from the standard space to the vector space by sending the $i$-th standard vector to the $i$-th basis vector of the given basis. This defines a unique linear mapping due to fact. Due to exercise, this mapping is bijective. It is just the mapping
    $(a_{1},\ldots ,a_{n})\longmapsto \sum _{i=1}^{n}a_{i}v_{i}.$
The inverse mapping
    $x=\Psi _{\mathfrak {v}}^{-1}\colon V\longrightarrow K^{n}$
is also linear, and it is called the coordinate mapping for this basis. The $i$-th component of this map, that is, the composed mapping
    $x_{i}=p_{i}\circ x\colon V\longrightarrow K,v\longmapsto (\Psi _{\mathfrak {v}}^{-1}(v))_{i},$
is called the $i$-th coordinate function. It is denoted by $v_{i}^{*}$. It assigns to vector $v\in V$ with the unique representation
    $v=\sum _{i=1}^{n}\lambda _{i}v_{i}\,$
the coordinate $\lambda _{i}$. Note that the linear mapping $v_{i}^{*}$ depends on the basis, not only on the vector $v_{i}$.
If an isomorphism
    $\Psi \colon K^{n}\longrightarrow V$
is given, then the images
    $\Psi (e_{i}),i=1,\ldots ,n,$
form a basis of $V$.
    Footnotes
* ↑ If $I$ is an infinite index set, then, in all sums considered here, only finitely many coefficients are not $0$.