# Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 48

    Commensurability

## Definition

Two line segments $s$ und $t$
$g$
$g$

## Definition

Real numbers $r,s\neq 0$ are called commensurable if ${\frac {r}{s}}$ is a
This is the equivalence relation for the subgroup
    $(\mathbb {Q} ^{\times },1,\cdot )\subseteq (\mathbb {R} ^{\times },1,\cdot )\,$
in the sense of definition.

## Remark

The question, whether there exist, beside the rational numbers, further useful numbers, has its roots in ancient Greece. The question was raised in the form whether two lengths, given in a natural way, are always commensurable to each other (whether there exists a third length such that the two given lengths are integer multiples of it). The Pythagorean school believed in the harmony of the universe, and this implied that all proportions between natural lengths must be expressed by a proportion of integer numbers. They found such integer proportions in music (music intervals), and thought that such proportions hold for the motions of the planets, and for geometry in general. Some people speculate that they knew the reasoning of Fact ***** showing the irrationality of ${\sqrt {2}}$ (the incommensurability of the side length and the length of the diagonal in a square), but that they concealed this knowledge. Anyhow, later in ancient times the knowledge was accepted that there are irrational numbers.
The subgroup relation
    $(\mathbb {Q} ,0,+)\subseteq (\mathbb {R} ,0,+)\,$
(which might be considered as a relation of a linear subspace of $\mathbb {Q}$-vector spaces) yields also an equivalence relation on the real numbers. Here, two real numbers are equivalent if their difference is a rational number.
    Residue class space

## Lemma
### Proof

$\Box$
We give another direct proof that we have an a equivalence relation.
Vector space/Linear subspace/Translation/Equivalence relation/Fact/Proof
The cosets of a linear subspace $U$ have a simple geometric interpretation: a coset is nothing but an affine subspace of $V$ that is parallel to $U$, that is, a space of the form $P+U$ with $P\in V$. The quotient set is the set of these affine subspaces.
We may apply to this equivalence relation the general results for a normal subgroup in a group and the corresponding equivalence relation, and we obtain a surjective quotient mapping (or identifying mapping, or canonical projection)
    $q\colon V\longrightarrow V/\sim ,v\longmapsto q(v)=[v]=v+U.$
instead of $V/\sim$, we shall write $V/U$. An extra feature in this situation is that this quotient set is itself a vector space, and that the canonical mapping is linear.

## Theorem
### Proof

$\Box$

## Definition

Vector space/Residue class space/Representative/Definition

## Theorem

Let $K$ be a field, and let $V,Q,$ and $W$ denote vector spaces over $K$. Let $\varphi \colon V\rightarrow W$ be a linear mapping, and $\psi \colon V\rightarrow Q$ be a surjective linear mapping. Suppose that
    $\operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,$
holds. Then there exists a uniquely determined linear mapping
    ${\tilde {\varphi }}\colon Q\longrightarrow W$
fulfilling $\varphi ={\tilde {\varphi }}\circ \psi$. In other words: the diagram
    ${\begin{matrix}V&{\stackrel {\varphi }{\longrightarrow }}&W&\\\!\!\!\!\!\,\,\psi \downarrow &\nearrow {\tilde {\varphi }}\!\!\!\!\!&\\Q&&&&\!\!\!\!\!\!\!\!\\\end{matrix}}$

### Proof

For every element $u\in Q$, there exists some $v\in V$ with $\psi (v)=u$. Due to the commutativity condition, we must have ${\tilde {\varphi }}(u)=\varphi (v)$. This means that there exists at most one ${\tilde {\varphi }}$. We have to show that, by this condition, a well-defined mapping is given. So let $v,v'\in V$ denote two preimages of $u$. Then
    $v'-v\in \operatorname {kern} \psi \subseteq \operatorname {kern} \varphi \,;$
therefore, $\varphi (v)=\varphi (v')$. hence, the mapping is well-defined. Let $u,u'\in Q$ be gievn, with preimages $v,v'\in V$. Then $v+v'$ is a preimage of $u+u'$; therefore, we have
    ${\tilde {\varphi }}(u+u')=\varphi (v+v')=\varphi (v)+\varphi (v')={\tilde {\varphi }}(u)+{\tilde {\varphi }}(u')\,.$
This means that ${\tilde {\varphi }}$ is compatible with the addition. Let $u\in Q$ be given with a preimage $v\in V$, and let $\lambda \in K$. Then $\lambda v$ is a preimage of $\lambda u$; therefore,
    ${\tilde {\varphi }}(\lambda u)=\varphi (\lambda v)=\lambda \varphi (v)=\lambda {\tilde {\varphi }}(u)\,,$
and ${\tilde {\varphi }}$ is also compatible with the scalar multiplication.
$\Box$
The mapping ${\tilde {\varphi }}$ constructed in the preceding theorem is called the induced linear mapping; accordingly, the theorem is also called the theorem about the induced mapping.

## Corollary

Let $K$ be a field, and let
    $\varphi \colon V\longrightarrow W$
denote a surjective linear mapping. Then there exists a canonical linear isomorphism
    ${\tilde {\varphi }}\colon V/\operatorname {kern} \varphi \longrightarrow W.$

### Proof

$\Box$

## Theorem
### Proof

$\Box$
This statement can be briefly formuluated as:
Image $=$ preimage modulo kernel.

## Lemma
### Proof

$\Box$

## Lemma
### Proof

$\Box$