Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 3

Groups

In linear algebra, we work in general over a fixed base field ${}K$. The most important field for us is the field of real numbers ${}\mathbb {R}$, which we have already used and which is introduced in analysis in an axiomatic way. Like the real numbers, a field is characterized by the existence of two binary operations fulfilling certain properties, namely addition and multiplication. Both these operations (for multiplication, one has to remove ${}0$) are instances of an important algebraic structure: groups.

Definition

A set ${}G$ together with a special element ${}e\in G$ and with a binary operation

$G\times G\longrightarrow G,(g,h)\longmapsto g\circ h,$

is called a group, when the following properties are fulfilled.

1. The binary operation is associative, i.e. for all ${}f,g,h\in G$ we have

   ${}(f\circ g)\circ h=f\circ (g\circ h)\,.$

2. The element ${}e$ is a neutral element, i.e. for all ${}g\in G$ we have

   ${}g\circ e=g=e\circ g\,.$

3. For every ${}g\in G$ there exists an inverse element, i.e. there exists some ${}h\in G$ such that

   ${}h\circ g=g\circ h=e\,.$

A group is called commutative, if the operation is commutative. Important examples for commutative groups are ${}(\mathbb {Z} ,0,+)$, ${}(\mathbb {R} ,0,+)$, ${}(\mathbb {R} \setminus \{0\},1,\cdot )$ or ${}(\mathbb {R} ^{n},0,+)$ with the componentewise zero

${}0=(0,0,\ldots ,0)\,$

and componentwise addition.

In a group ${}(G,e,\circ )$, the neutral element is uniquely determined. For if ${}e'$ is another element fulfilling the characteristic property of the neutral element, meaning

${}x\circ e'=e'\circ x=x\,$

for all ${}x\in G$, then we can deduce directly

${}e=e\circ e'=e'\,.$

Lemma

Let ${}(G,e,\circ )$ be a group. Then for every ${}x\in G$, the element ${}y\in G$ fulfilling

${}x\circ y=y\circ x=e\,$

is uniquely determined.

Proof

Let

${}x\circ y=y\circ x=e\,$

and

${}x\circ z=z\circ x=e\,.$

Then we have

${}y=y\circ e=y\circ (x\circ z)=(y\circ x)\circ z=e\circ z=z\,.$

$\Box$

Abstract structures like a set, a mapping, a binary operation have a double life: On one hand, they are really just the given formal structure, the elements are just some elements in a somehow given set, a binary operation is just any binary operation, and one should not imagine anything concrete. The symbols chosen are arbitrary and without any meaning. On the other hand, these abstract structures gain a second life in that many concrete mathematical structures obey the abstract properties. These concrete structures are examples or models for the abstract structure (and they are also motivation to introduce the abstract structure). Both viewpoints are important, and one should always try not to confuse them.

Group theory is a branch of mathematics on is own, which we will not develop here systematically. Instead, we work with rings and in particular with fields.

Rings

Definition

A set ${}R$ is called a ring if there are two binary operations (called addition and multiplication)

$+:R\times R\longrightarrow R{\text{ and }}\cdot :R\times R\longrightarrow R$

and two elements ${}0,1\in R$, which fulfill the following properties.

1. Axioms for the addition:
   1. Law of associativity: ${}(a+b)+c=a+(b+c)$ holds for all ${}a,b,c\in R$.
   2. Law of commutativity: ${}a+b=b+a$ holds for all ${}a,b\in R$.
   3. ${}0$ is the neutral element of the addition, i.e. ${}a+0=a$ holds for all ${}a\in R$.
   4. Existence of the negative: For every ${}a\in R$, there exists an element ${}b\in R$ with ${}a+b=0$.
2. Axioms of the multiplication:
   1. Law of associativity: ${}(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all ${}a,b,c\in R$.
   2. ${}1$ is the neutral element for the multiplication, i.e. ${}a\cdot 1=a$ holds for all ${}a\in R$.
3. Law of distributivity: ${}a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all ${}a,b,c\in R$.

Definition

A ring is called commutative

it its multiplication is commutative.

For us, the most important commutative rings are the set of integer numbers ${}\mathbb {Z}$, the rational numbers ${}\mathbb {Q}$ and the real numbers ${}\mathbb {R}$. The real numbers (and the rational numbers) with their natural operations fulfill all these axioms, as should be known from school. An axiomatic reasoning is possible, but we will not do this here. With its addition, a ring ${}(R,0,+)$ (forgetting the multiplicative structure) is in particular a commutative group.

In a Ring, we use the convention that multiplication ties stronger than addition. Therefore, we write ${}a\cdot b+c\cdot d$ instead of ${}(a\cdot b)+(c\cdot d)$. To simplify further the notation, we omit the product symbol. The special elements ${}0$ and ${}1$ in a ring are called the null element and the unit. For ${}a\in R$, we call the (according to) fact uniquely determined element ${}y$ fulfilling ${}a+y=0$ the negative of ${}a$ and denote it by ${}-a$. We have ${}-(-a)=a$, since ${}a+(-a)=0$ shows that the element ${}a$ equals the uniquely determined negative of ${}-a$. We write ${}b-a$ instead of ${}b+(-a)$ and call this a difference. Hence, the difference is not a basic operation, but is defined as the addition with the negative element.

The following properties are familiar for the real numbers, we prove them using only the axioms of a rings. So they hold for any ring.

Lemma

Let ${}R$ be a ring,

and let ${}a,b,c,a_{1},\ldots ,a_{r},b_{1},\ldots ,b_{s}$ denote elements from ${}R$. Then the following statements hold.

1. ${}a0=0$ (annulation rule).

2. ${}(-a)b=-ab=a(-b)\,$

   (rules for sign).

3. ${}(-a)(-b)=ab\,.$

4. ${}a(b-c)=ab-ac\,.$

5. ${}{\left(\sum _{i=1}^{r}a_{i}\right)}{\left(\sum _{k=1}^{s}b_{k}\right)}=\sum _{1\leq i\leq r,\,1\leq k\leq s}a_{i}b_{k}$ (general law of distributivity).

Proof

In the noncommutative case we only proof one half of the statements.

1. We have ${}a0=a(0+0)=a0+a0$. Subtracting ${}a0$ (meaning addition with the negative of ${}a0$) on both sides gives the claim.

2. ${}(-a)b+ab=(-a+a)b=0b=0\,$

   due to part (1). Therefore ${}(-a)b$ is the (uniquely determined) negative of ${}ab$.

3. Due to (2), we have ${}(-a)(-b)=(-(-a))b$ and because of ${}-(-a)=a$ (which holds in every group) we get the claim.
4. This follows from the parts proved so far.
5. This follows with a double induction

$\Box$

Fields

A large part of linear algebra might be worked out over an arbitrary commutative ring, but that needs many more additional concepts. In this course, we will work over a field.

Definition

A commutative ring ${}K$ is called a field if ${}R\neq 0$

and if every element different from ${}0$ has a multiplicative inverse.

In all details, this means the following.

Definition

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.

1. Axioms for the addition:
   1. Law of associativity: ${}(a+b)+c=a+(b+c)$ holds for all ${}a,b,c\in K$.
   2. Law of commutativity: ${}a+b=b+a$ holds for all ${}a,b\in K$.
   3. ${}0$ is the neutral element of the addition, i.e. ${}a+0=a$ holds for all ${}a\in K$.
   4. Existence of the negative: For every ${}a\in K$, there exists an element ${}b\in K$ with ${}a+b=0$.
2. Axioms of the multiplication:
   1. Law of associativity: ${}(a\cdot b)\cdot c=a\cdot (b\cdot c)$ holds for all ${}a,b,c\in K$.
   2. Law of commutativity: ${}a\cdot b=b\cdot a$ holds for all ${}a,b\in K$.
   3. ${}1$ is the neutral element for the multiplication, i.e. ${}a\cdot 1=a$ holds for all ${}a\in K$.
   4. Existence of the inverse: For every ${}a\in K$ with ${}a\neq 0$, there exists an element ${}c\in K$ such that ${}a\cdot c=1$.
3. Law of distributivity: ${}a\cdot (b+c)=(a\cdot b)+(a\cdot c)$ holds for all ${}a,b,c\in K$.

The properties described in fact for rings (and the conventions) hold in particular for fields. Using the concept of a group, we may say that a field is a set with two binary operations Verknüpfungen ${}+$ and ${}\cdot$ and two fixed elements ${}0\neq 1$, such that ${}(K,+,0)$ and ${}(K\setminus \{0\},\cdot ,1)$ are commutative groups and that the distributivity law holds.

For an element ${}x\in K$ and a natural number ${}n\in \mathbb {N}$ we define ${}nx$ to be the ${}n$-fold sum of ${}x$ with itself. Here we put ${}0x=0$. For

${}n1_{K}=\underbrace {1_{K}+1_{K}+\cdots +1_{K}} _{n{\text{ Summanden}}}\,,$

we also write simply ${}n_{K}$ or just ${}n$. This means that we can find every natural number in every field (also in every ring). However, this assignemnt is not necessarily injective, and it is possible that ${}2=0$ or ${}7=0$ holds in a field (see the examples below). For a negative integer ${}n$ we set

${}nx=(-n)(-x)\,,$

where ${}-x$ denotes the negative of ${}x$ in the field. Due to exercise, everything fits well together. For example, one may consider ${}(-n)(-x)$ as the ${}-n$-fold sum of ${}-x$ with itself, or as the product of ${}-x$ and ${}-n$, where this means the ${}-n$-fold sum of ${}1_{K}$ with itself.

The graph of the real function which assigns for a number ${}a\neq 0$ its inverse. This mapping is not defined in ${}0$, nor can it be extended in a continuous way.

Due to fact we know that for every ${}a\in K$, ${}a\neq 0$, the element ${}z$ fulfilling ${}az=1$ is unique. It is called the inverse of ${}a$ and denoted by ${}a^{-1}$.

For ${}a,b\in K$, ${}b\neq 0$, we write

${}a/b:={\frac {a}{b}}:=ab^{-1}\,.$

The terms on the left are abbreviations for the term on the right.

For a field element ${}a\in K$ and ${}n\in \mathbb {N}$ we denote its ${}n$-th power by ${}a^{n}$, this is defined as the ${}n$-fold product of ${}a$ with itself (${}n$ is the number of factors). Moreover, we set ${}a^{0}=1$, and, for ${}a\neq 0$ and ${}n\in \mathbb {N} _{+}$, we interpret ${}a^{-n}$ as ${}{\left(a^{-1}\right)}^{n}$.

A "strange“ field is given in the following example. This field with two elements is important in computer science and in coding theory, it will not play a big role here. It shows that it is not for every field helpful to imagine its elements on the number line.

Example

We are trying to find a structure of a field on the set ${}\{0,1\}$. If ${}0$ is supposed to be the neutral element of the addition and ${}1$ the neutral element of the multiplication, then everything is already determined: The equation ${}1+1=0$ must hold, since ${}1$ has an inverse element with respect to the addition, and since ${}0\cdot 0=0$ holds, due to fact. Hence the operation tables look like

${}+$	${}0$	${}1$
${}0$	${}0$	${}1$
${}1$	${}1$	${}0$

and

${}\cdot$	${}0$	${}1$
${}0$	${}0$	${}0$
${}1$	${}0$	${}1$

With some tedious computations, one can check that this is indeed a field.

Example

On the set ${}\{0,1,2,3,4,5,6\}$ (with seven elements) one can define a field structure using

${}+$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}0$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}1$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$	${}0$
${}2$	${}2$	${}3$	${}4$	${}5$	${}6$	${}0$	${}1$
${}3$	${}3$	${}4$	${}5$	${}6$	${}0$	${}1$	${}2$
${}4$	${}4$	${}5$	${}6$	${}0$	${}1$	${}2$	${}3$
${}5$	${}5$	${}6$	${}0$	${}1$	${}2$	${}3$	${}4$
${}6$	${}6$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$

${}\cdot$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}0$	${}0$	${}0$	${}0$	${}0$	${}0$	${}0$	${}0$
${}1$	${}0$	${}1$	${}2$	${}3$	${}4$	${}5$	${}6$
${}2$	${}0$	${}2$	${}4$	${}6$	${}1$	${}3$	${}5$
${}3$	${}0$	${}3$	${}6$	${}2$	${}5$	${}1$	${}4$
${}4$	${}0$	${}4$	${}1$	${}5$	${}2$	${}6$	${}3$
${}5$	${}0$	${}5$	${}3$	${}1$	${}6$	${}4$	${}2$
${}6$	${}0$	${}6$	${}5$	${}4$	${}3$	${}2$	${}1$

Without any further theory, it is very tedious to so show that this is indeed a field.

Lemma

Let ${}K$ denote a field. Then ${}a\cdot b=0$ implies that ${}a=0$ or ${}b=0$.

Proof

We prove this by contradiction, so we assume that ${}a$ and ${}b$ are both not ${}0$. Then there exist inverse elements ${}a^{-1}$ and ${}b^{-1}$ and hence ${}(ab){\left(b^{-1}a^{-1}\right)}=1$. On the other hand, we have ${}ab=0$ by the premise and so the annulation rule gives

${}(ab){\left(b^{-1}a^{-1}\right)}=0{\left(b^{-1}a^{-1}\right)}=0\,,$

hence ${}0=1$, which contradicts the field properties.

$\Box$

Footnotes

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