# Linear algebra (Osnabrück 2024-2025)/Part I/List of definitions

The set which does not contain any element is called the empty set, denoted by
    $\emptyset .$
Let $T$ and $M$ denote sets. $T$ is called a subset
$M$
$T$
$M$
For sets $L$ und $M$, we call
    $L\cap M={\left\{x\mid x\in L{\text{ and }}x\in M\right\}}\,$
the intersection
For sets $L$ und $M$, we call
    $L\cup M={\left\{x\mid x\in L{\text{ or }}x\in M\right\}}\,$
the union
Suppose that two sets $L$ and $M$ are given. Then the set
    $L\times M={\left\{(x,y)\mid x\in L,\,y\in M\right\}}\,$
is called the product set
Let $L$ and $M$ denote sets. A mapping $F$ from $L$ to $M$ is given by assigning, to every element of the set $L$, exactly one element of the set $M$. The unique element which is assigned to $x\in L$, is denoted by $F(x)$. For the mapping as a whole, we write
    $F\colon L\longrightarrow M,x\longmapsto F(x).$
Let $L$ and $M$ denote sets, and let
    $F\colon L\longrightarrow M,x\longmapsto F(x),$
be a mapping. Then $F$ is called injective, if for two different elements $x,x'\in L$, also $F(x)$ and $F(x')$
Let $L$ and $M$ denote sets, and let
    $F\colon L\longrightarrow M,x\longmapsto F(x),$
be a mapping. Then $F$ is called surjective, if for every $y\in M$, there exists at least one element $x\in L$, such that
    $F(x)=y\,.$
Let $M$ and $L$ denote sets and suppose that
    $F\colon M\longrightarrow L,x\longmapsto F(x),$
is a mapping. Then $F$ is called bijective if $F$ is injective as well as
Let $F\colon L\rightarrow M$ denote a bijective mapping. Then the mapping
    $G\colon M\longrightarrow L$
which sends every element $y\in M$ to the uniquely determined element $x\in L$ with $F(x)=y$,
$F$
Let $L,\,M$ and $N$ denote sets, let
    $F\colon L\longrightarrow M,x\longmapsto F(x),$
and
    $G\colon M\longrightarrow N,y\longmapsto G(y),$
be mappings. Then the mapping
    $G\circ F\colon L\longrightarrow N,x\longmapsto G(F(x)),$
is called the composition of the mappings
$F$
$G$
An operation (or binary operation) $\circ$ on a set $M$ is a mapping
    $\circ \colon M\times M\longrightarrow M,(x,y)\longmapsto \circ (x,y)=x\circ y.$
A set $K$ is called a field if there are two binary operations (called addition and multiplication)
    $+:K\times K\longrightarrow K{\text{ and }}\cdot :K\times K\longrightarrow K$
and two different elements $0,1\in K$, which fulfill the following properties.
* Axioms for the addition:
** Law of associativity: $(a+b)+c=a+(b+c)$ holds for all $a,b,c\in K$.
** Law of commutativity: $a+b=b+a$ holds for all $a,b\in K$.
** $0$ is the neutral element of the addition, i.e. $a+0=a$ holds for all $a\in K$.
** Existence of the negative: For every $a\in K$, there exists an element $b\in K$ with $a+b=0$.
* Axioms of the multiplication:
** Law of associativity: $(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all $a,b,c\in K$.
** Law of commutativity: $a\cdot b=b\cdot a$ holds for all $a,b\in K$.
** $1$ is the neutral element for the multiplication, i.e. $a\cdot 1=a$ holds for all $a\in K$.
** Existence of the inverse: For every $a\in K$ with $a\neq 0$, there exists an element $c\in K$ such that $a\cdot c=1$.
* Law of distributivity: $a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all $a,b,c\in K$.
For a natural number $n$, one puts
    $n!:=n(n-1)(n-2)\cdots 3\cdot 2\cdot 1\,,$
$n$
Let $k$ and $n$ denote natural numbers with $k\leq n$. Then
    ${\binom {n}{k}}:={\frac {n!}{k!(n-k)!}}\,$
$n$
$k$
Let $K$ be a field. An expression of the form
    $P=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,,$
with $a_{i}\in K$ and $n\in \mathbb {N}$,
$K$
The degree of a nonzero polynomial
    $P=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,$
with $a_{n}\neq 0$
$n$