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

    Commensurability

## Definition

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

## Definition

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

Let $K$ be a field,MDLD/field let $V$ be a $K$-vector space,MDLD/vector space and let $U\subseteq V$ denote a linear subspace.MDLD/linear subspace Then the relationMDLD/relation given by
    $v\sim w,{\text{ falls }}v-w\in U,$
is an equivalence relationMDLD/equivalence relation
$V$

### Proof

$\Box$
We give another direct proof that we have an a equivalence relation.
We look at the conditions of an equivalence relationnMDLD/equivalence relation The reflexivity follows from $v-v=0\in U$, the symmetry follows from $w-v=-(v-w)\in U$. The transitivity follows like this: from $u-v\in U$ and $v-w\in U$, we infer $u-w=(u-v)+(v-w)\in U$.
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.

## TheoremTheorem 48.5 change

Let $K$ be a field,MDLD/field let $V$ be a $K$-vector space,MDLD/vector space and let $U\subseteq V$ denote a linear subspace.MDLD/linear subspace Let $V/U$ be the set of all equivalence classesMDLD/equivalence classes (the quotient set) of the equivalence relationMDLD/equivalence relation (linear subspace) on $V$ defined by $U$, and let
    $q\colon V\longrightarrow V/U,v\longmapsto [v],$
denote the canonical projection.MDLD/canonical projection Then there exists a uniquely determined structure of a $K$-vector space on $V/U$ such that $q$ is a
$K$

### Proof

The existence and uniqueness of the group structure on $V/U$, and the homomorphism property of the projection, follow from Theorem 46.14 . For the projection to be linear, the scalar multiplication must satisfy
    $\lambda [v]=[\lambda v]\,.$
We have to show that by this condition, a well-defined scalar multiplication on $V/U$ is given. Let $v'=v+u$, with $u\in U$. Then we have
    $\lambda v'=\lambda (v+u)=\lambda v+\lambda u\,,$
and the is equivalent to $\lambda v$. The well-definedness of the scalar multiplication on $V/U$ and the surjectivity of the mapping imply that we have a vector space structure, and that the mapping is also compatible with the scalar multiplication.
$\Box$

## Definition

Let $K$ be a field,MDLD/field let $V$ be a $K$-vector space,MDLD/vector space and let $U\subseteq V$ denote a linear subspace.MDLD/linear subspace Then the set $V/U$ of all equivalence classes,MDLD/equivalence classes together with the structure of a vector space as proven in Theorem 48.5 , is called the residue class space (or quotient space)
$V$
$U$

## TheoremTheorem 48.7 change

Let $K$ be a field,MDLD/field and let $V,Q,$ and $W$ denote vector spacesMDLD/vector spaces over $K$. Let $\varphi \colon V\rightarrow W$ be a linear mapping,MDLD/linear mapping and $\psi \colon V\rightarrow Q$ be a surjectiveMDLD/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

The statement about existence and uniqueness follows from Theorem 47.1 . We have to show that the uniquly determined group homomorphism ${\tilde {\varphi }}$ is also compatible with scalar multiplication. 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.

## CorollaryCorollary 48.8 change

Let $K$ be a field,MDLD/field and let
    $\varphi \colon V\longrightarrow W$
denote a surjectiveMDLD/surjective linear mappingMDLD/linear mapping between $K$-vector spaces.MDLD/vector spaces Then there exists a canonical linear isomorphismMDLD/linear isomorphism
    ${\tilde {\varphi }}\colon V/\operatorname {kern} \varphi \longrightarrow W.$

### Proof

This follows from Corollary 47.3
and Theorem 48.7 .
$\Box$

## TheoremCreate referencenumber

Let $K$ be a field,MDLD/field and let
    $\varphi \colon V\longrightarrow W$
be a linear mappingMDLD/linear mapping between $K$-vector spaces.MDLD/vector spaces Then there exists a canonical factorization
    $V{\stackrel {q}{\longrightarrow }}V/\operatorname {kern} \varphi {\stackrel {\theta }{\longrightarrow }}\operatorname {Im} \varphi {\stackrel {\iota }{\hookrightarrow }}W,$
where $q$ is the canonical projection,MDLD/canonical projection $\theta$ is a vector space isomorphism,MDLD/vector space isomorphism and $i$ is the canonical inclusion of the image spaceMDLD/image space
$W$

### Proof

This follows from Corollary 48.8 , applied to the surjective linear mapping
    $V\longrightarrow \operatorname {Im} \varphi .$
$\Box$
This statement can be briefly formuluated as:
Image $=$ preimage modulo kernel.

## LemmaCreate referencenumber

Let $K$ denote a field,MDLD/field and let $V$ denote a finite-dimensionalMDLD/finite-dimensional (fgvs) vector space.MDLD/vector space Let $U\subseteq V$ be a linear subspace.MDLD/linear subspace Then
    $\dim _{K}{\left(V/U\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,.$

### Proof

The canonical projectionMDLD/canonical projection
    $V\longrightarrow V/U$
is surjective;MDLD/surjective its kernel is $U$. Because of the dimension formula, we obtain
    $\dim _{K}{\left(V/U\right)}=\dim _{K}{\left(V\right)}-\dim _{K}{\left(U\right)}\,.$
$\Box$

## LemmaCreate referencenumber

Let $K$ be a field,MDLD/field and let $V$ denote a $K$-vector space,MDLD/vector space together with a direct sum decompositionMDLD/direct sum decomposition
    $V=U\oplus W\,,$
with linear subspacesMDLD/linear subspaces $U$ and $W$. Then
    $V/U\cong W\,.$

### Proof

The kernelMDLD/kernel of the projection
    $V\longrightarrow W$
is $U$. Therefore, the statement follows from Corollary 48.8 .
$\Box$