OpenLLM-France/wikimedia huggingface dataset:
# Endomorphism/Geometric multiplicity/Section

The restriction of a linear mapping to an eigenspace is the homothety with the corresponding eigenvalue, thus a very simple linear mapping. For a diagonal matrix
    ${\begin{pmatrix}d_{1}&0&\cdots &\cdots &0\\0&d_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&d_{n-1}&0\\0&\cdots &\cdots &0&d_{n}\end{pmatrix}},$
the standard basis has the property that every basis vector is an eigenvector for the linear mapping given by the matrix. In this case, it is easy to describe the eigenspaces, see example; the eigenspace for $d$ consists of all linear combinations of the standard vectors $e_{i}$, for which $d_{i}$ equals $d$. In particular, the dimension of the eigenspace equals the number how often $d$ occurs as an diagonal element. In general, the dimensions of the eigenspaces are important invariants of an endomorphism.

## Definition

Endomorphism/Eigenvalue/Geometric multiplicity/Definition
In particular, a number $\lambda \in K$ is an eigenvalue of $\varphi$ if and only if its geometric multiplicity is at least $1$. It is easy to give examples where the geometric multiplicity of an eigenvalue is any number between $1$ and the dimension of the space.

## Lemma
### Proof

$\Box$
# Endomorphism/Eigenvalue/Geometric multiplicity/Definition

Let $K$ be a field, $V$ a $K$-vector space and
    $\varphi \colon V\longrightarrow V$
a linear mapping. For $\lambda \in K$ we call
    $\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$
the geometric multiplicity of $\lambda$.
# Eigenspaces/Direct sum/Dimension/Fact

Let $K$ denote a field, and let $V$ denote a $K$-vector space of finite dimension. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping.
sum of the eigenspaces $\operatorname {Eig} _{\lambda }{\left(\varphi \right)}$ is direct, and we have
    $\sum _{\lambda \in K}\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \dim _{\left(V\right)}\,.$
# Matrix/2x2/Shearing matrices/Eigenvalues and diagonalizable/Example

Matrix/2x2/Shearing matrices/Eigenvalues/Example For $a\neq 0$, there is a one-dimensional eigenspace, and the mapping is not diagonalizable.
# Matrix/2x2/Shearing matrices/Eigenvalues/Example

We consider $2\times 2$-shearing matrices
    ${\begin{pmatrix}1&a\\0&1\end{pmatrix}}$
with $a\in K$. The condition for some $\lambda \in K$ to be an eigenvalue means
    ${\begin{pmatrix}1&a\\0&1\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}=\lambda {\begin{pmatrix}x\\y\end{pmatrix}}\,.$
This yields the equations
    $x+ay=\lambda x{\text{ and }}y=\lambda y.$
For $\lambda \neq 1$, we get $y=0$ and hence also $x=0$, that is, only $1$ can be an eigenvalue. In this case, the second equation is fulfilled, and the first equation becomes
    $x+ay=x{\text{ and }}ay=0.$
For $a\neq 0$, we get $y=0$ and thus ${\begin{pmatrix}x\\0\end{pmatrix}}$ is the eigenspace for the eigenvalue $1$, and ${\begin{pmatrix}1\\0\end{pmatrix}}$ is an eigenvector which spans this eigenspace. For $a=0$, we have the identity matrix, and the eigenspace for the eigenvalue $1$ is the total plane.
# Endomorphism/Multiplicities/Diagonalizability/Section
## Theorem

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping. Then $\varphi$ is diagonalizable if and only if the characteristic polynomial $\chi _{\varphi }$ is a product of linear factors and if for every zero $\lambda$ with algebraic multiplicity $\mu _{\lambda }$, the identity
    $\mu _{\lambda }=\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\,$

### Proof

$\Box$

## Corollary
### Proof

$\Box$
This gives also a new proof for fact.
# Endomorphism/Eigenvalue/Algebraic multiplicity/Definition

Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping on a finite-dimensional $K$-vector space $V$, and $\lambda \in K$. The exponent of the linear polynomial $X-\lambda$ in the characteristic polynomial $\chi _{\varphi }$ is called the algebraic multiplicity of $\lambda$. It is denoted by
    $\mu _{\lambda }=\mu _{\lambda }(\varphi )\,.$
# Endomorphism/Characteristic polynomial/Different linear factors/Diagonalizable/Fact

Let $K$ denote a field, and let $V$ denote a $K$-vector space of finite dimension. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping. Suppose that the characteristic polynomial $\chi _{\varphi }$ factors into different linear factors.
$\varphi$
# Endomorphism/Algebraic multiplicity/Introduction/Section

For a more detailed investigation of the eigenspaces, the following concept is useful.

## Definition

Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping on a finite-dimensional $K$-vector space $V$, and $\lambda \in K$. The exponent of the linear polynomial $X-\lambda$ in the characteristic polynomial $\chi _{\varphi }$ is called the algebraic multiplicity of $\lambda$. It is denoted by
    $\mu _{\lambda }=\mu _{\lambda }(\varphi )\,.$
Recall that
    $\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$
is called the geometric multiplicity of $\lambda$. We know, due to fact, that one of these multiplicities is positive if and only if this is true for the other multiplicity, and this is the case if and only if $\lambda$ is an eigenvalue.
In general, the multiplicities can be different, we have, however, an estimate between them.

## Lemma

Let $K$ denote a field, and let $V$ denote a finite-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
denote a linear mapping and $\lambda \in K$. Then we have the estimate
    $\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}\leq \mu _{\lambda }(\varphi )\,$
between the geometric and the

### Proof

Let $m=\dim _{\left(\operatorname {Eig} _{\lambda }{\left(\varphi \right)}\right)}$ and let $v_{1},\ldots ,v_{m}$ be a basis of this eigenspace. We complement this basis with $w_{1},\ldots ,w_{n-m}$ to get a basis of $V$, using fact. With respect to this basis, the describing matrix has the form
    ${\begin{pmatrix}\lambda E_{m}&B\\0&C\end{pmatrix}}.$
The characteristic polynomial equals therefore (using exercise) $(X-\lambda )^{m}\cdot \chi _{C}$, so that the algebraic multiplicity is at least $m$.
$\Box$

## Example

We consider the $2\times 2$-shearing matrix
    $M={\begin{pmatrix}1&a\\0&1\end{pmatrix}}\,,$
with $a\in K$. The characteristic polynomial is
    $\chi _{M}=(X-1)(X-1)\,,$
so that $1$ is the only eigenvalue of $M$. The corresponding eigenspace is
    $\operatorname {Eig} _{1}{\left(M\right)}=\operatorname {ker} {\left({\begin{pmatrix}0&-a\\0&0\end{pmatrix}}\right)}\,.$
From
    ${\begin{pmatrix}0&-a\\0&0\end{pmatrix}}{\begin{pmatrix}r\\s\end{pmatrix}}={\begin{pmatrix}-as\\0\end{pmatrix}}\,,$
we get that ${\begin{pmatrix}1\\0\end{pmatrix}}$ is an eigenvector, and in case $a\neq 0$, the eigenspace is one-dimensional (in case $a=0$, we have the identity and the eigenspace is two-dimensional). So in case $a\neq 0$, the algebraic multiplicity of the eigenvalue $1$ equals $2$, and the geometric multiplicity equals $1$.
# Endomorphism/Minimal polynomial/Characteristic polynomial/Section
## Corollary
### Proof

$\Box$
In particular, the degree of the minimal polynomial of $\varphi \colon V\rightarrow V$ is bounded by the dimension of the vector space $V$. The minimal polynomial and the characteristic polynomial are related in several respects, for example, they have the same zeroes.

## Lemma
### Proof

$\Box$

## Lemma
### Proof

$\Box$
# Roots of unity/Field/Definition

Let $K$ be a field, and $n\in \mathbb {N}$ A zero of the polynomial
    $X^{n}-1$
in $K$ is called an $n$-th root of unity in $K$.
# Wikiphilosophers/Fate

Fate is determined, there is no way that the universe could not be deterministic, it may appear random in some ways to us. However, without using religious ideas to justify randomness, or something that seems random because we can't see the dimension in which it is deterministic. Nothing cannot come from something, therefore everything is deterministic, and if not it is controlled by a god, a god which is timeless and not limited by the physical realm.
# Endomorphism/Trigonalizable/Characterizations/Section
## Definition

Vector space/Flag/Definition
Hence, a flag is a chain of linear subspaces, contained in each other, and where the dimension increase by $1$ in each step.

## Definition

Linear mapping/Invariant flag/Definition

## Theorem
### Proof

$\Box$

## Theorem

Let $M\in \operatorname {Mat} _{n\times n}(\mathbb {C} )$ denote a square matrix with complex entries. Then $M$ is

### Proof

This follows from fact and the Fundamental theorem of algebra.
$\Box$
# Matrix/31-11/Trigonalizable/Similar/Example

We claim that the matrix
    $M={\begin{pmatrix}3&1\\-1&1\end{pmatrix}}\,$
is trigonalizable. The matrix
    $B={\begin{pmatrix}3&2\\1&1\end{pmatrix}}\,$
is invertible with the inverse matrix
    $B^{-1}={\begin{pmatrix}1&-2\\-1&3\end{pmatrix}}\,.$
A direct computation shows
    $BMB^{-1}={\begin{pmatrix}3&2\\1&1\end{pmatrix}}{\begin{pmatrix}3&1\\-1&1\end{pmatrix}}{\begin{pmatrix}1&-2\\-1&3\end{pmatrix}}={\begin{pmatrix}3&2\\1&1\end{pmatrix}}{\begin{pmatrix}2&-3\\-2&5\end{pmatrix}}={\begin{pmatrix}2&1\\0&2\end{pmatrix}}\,.$
In this verification of trigonalizability, the transformation matrix $B$ arises just like that. We get a more reasonable proof with the help of the characteristic polynomial and fact. The characteristic polynomial is
    $\chi _{M}=\det {\begin{pmatrix}X-3&-1\\1&X-1\end{pmatrix}}=(X-3)(X-1)+1=X^{2}-4X+4=(X-2)^{2}\,,$
and this factors into linear factors.
# Endomorphism/Invariant linear subspace/Polynomial/Action/Fact/Proof

This can be checked directly for the powers $X^{n}$ and for linear combinations of powers.
# Linear mapping/Invariant flag/Definition

Let $V$ be a vector space of dimension $n$, and let
    $f\colon V\longrightarrow V$
be a linear mapping. A flag
    $0=V_{0}\subset V_{1}\subset \ldots \subset V_{n-1}\subset V_{n}\,$
is called $f$-invariant, if $f(V_{i})\subseteq V_{i}$ holds for all $i=0,1,\ldots ,n-1,n$.
# Real matrix/2x2/Trigonalizability via characteristic polynomial/Example

We consider a real $2\times 2$-matrix $M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}$. The characteristic polynomial is
    ${\begin{aligned}\chi _{M}&=\det {\left(xE_{2}-M\right)}\\&=\det {\begin{pmatrix}x-a&-b\\-c&x-d\end{pmatrix}}\\&=(x-a)(x-d)-bc\\&=x^{2}-(a+d)x+ad-bc\\&={\left(x-{\frac {a+d}{2}}\right)}^{2}-{\left({\frac {a+d}{2}}\right)}^{2}+ad-bc\\&={\left(x-{\frac {a+d}{2}}\right)}^{2}-{\left({\frac {a-d}{2}}\right)}^{2}-bc.\end{aligned}}$
This polynomial splits into (real) linear factors if and only if ${\left({\frac {a-d}{2}}\right)}^{2}+bc\geq 0$. The matrix is trigonalizable exactly in this case, due to fact.
# Trigonalizable/Direct sum/Fact

Let $V_{1},\ldots ,V_{n}$ denote finite-dimensional vector spaces over the field $K$, let
    $\varphi _{i}\colon V_{i}\longrightarrow V_{i}$
denote linear mappings, and let
    $\varphi =\varphi _{1}\times \cdots \times \varphi _{n}\colon V_{1}\times \cdots \times V_{n}\longrightarrow V_{1}\times \cdots \times V_{n}$
denote the product mapping.
$\varphi$
trigonalizable
$\varphi _{i}$
# Trigonalizable/Direct sum/Fact/Proof/Exercise

Let $V_{1},\ldots ,V_{n}$ denote finite-dimensional vector spaces over the field $K$, let
    $\varphi _{i}\colon V_{i}\longrightarrow V_{i}$
denote linear mappings, and let
    $\varphi =\varphi _{1}\times \cdots \times \varphi _{n}\colon V_{1}\times \cdots \times V_{n}\longrightarrow V_{1}\times \cdots \times V_{n}$
denote the product mapping. Show that $\varphi$ is trigonalizable if and only if this holds for all $\varphi _{i}$.
# Endomorphism/Eigenvalue/Invariant hyperplane/Fact

Let $K$ denote a field, and let $V$ be a $n$-dimensional vector space. Let
    $\varphi \colon V\longrightarrow V$
be a linear mapping, and let $\lambda \in K$ be an eigenvalue of $\varphi$.
$\varphi$-invariant linear subspace $U\subset V$
$n-1$
# Endomorphism/Eigenvalue/Invariant hyperplane/Fact/Proof

Because of the condition and
fact, the mapping $\varphi -\lambda \operatorname {Id} _{V}$ has a nontrival kernel. Hence, this mapping is not injective and, due to fact, also not surjective. Therefore,
    $B:=\operatorname {Im} {\left(\varphi -\lambda \operatorname {Id} _{V}\right)}\subset V\,$
is a strict linear subspace of $V$. It follows that there exists also a linear subspace $U\subset V$ of dimension $n-1$, which contains $B$. For $u\in U$, we have
    $\varphi (u)=\lambda u+(\varphi -\lambda \operatorname {Id} _{V})u\in U+B\subseteq U\,.$
Hence, the image of $U$ belongs to $U$, that is, $U$ is $\varphi$-invariant.
# Cayley-Hamilton/Minimal polynomial and characteristic polynomial/Fact

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
be a linear mapping.
characteristic polynomial $\chi _{f}$ is a multiple of the minimal polynomial
$\mu _{f}$
$f$
# Cayley-Hamilton/Matrix version/Fact

Let $K$ be a field, and let $M$ be an $n\times n$-matrix. Let
    $\chi _{M}=X^{n}+c_{n-1}X^{n-1}+\cdots +c_{1}X+c_{0}\,$
denote the characteristic polynomial of $M$.
    $\chi _{M}\,(M)=M^{n}+c_{n-1}M^{n-1}+\cdots +c_{1}M+c_{0}=0\,.$
This means that the matrix annihilates the characteristic polynomial.
# Cayley-Hamilton/Fact

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
denote a linear mapping.
characteristic polynomial of $f$ fulfills the relation
    $\chi _{f}(f)=0\,.$
# Cayley-Hamilton/Fact/Proof

This follows immediately from fact.
# Cayley-Hamilton/Minimal polynomial and characteristic polynomial/Fact/Proof

This follows directly from fact and fact.
# Endomorphism/Polynomial/Eigenvector/Fact

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
be a linear mapping. Let $v\in V$ be an eigenvector of $f$ with eigenvalue $\lambda$, and let $P\in K[X]$ denote a polynomial.
    $(P(f))(v)=P(\lambda )v\,.$
In particular, $v$ is an eigenvector of $P(f)$ with eigenvalue $P(\lambda )$. The vector $v\neq 0$ belongs to the kernel of $P(f)$ if and only if $\lambda$ is a zero of $P$.
# Endomorphism/Polynomial/Eigenvector/Fact/Proof

We have
    $(f^{k})(v)=\lambda ^{k}v\,.$
This implies the statement, since the assignment $P\mapsto P(f)$ is compatible with addition and scalar multiplication.
# Minimal polynomial and characteristic polynomial/Same zeroes/Fact

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
be a linear mapping.
characteristic polynomial $\chi _{f}$ and the minimal polynomial
$\mu _{f}$
# Minimal polynomial and characteristic polynomial/Same zeroes/Fact/Proof

It follows directly from Cayley-Hamilton that the zeroes of the minimal polynomial are also zeroes of the characteristic polynomial.
To prove the other implication, let $\lambda \in K$ be a zero of the characteristic polynomial, and let $v\in V$ denote an eigenvector of $f$ with eigenvalue $\lambda$, its existence is guaranteed by fact. We write the minimal polynomial as
    $\mu _{f}=(X-\lambda _{1})^{m_{1}}\cdots (X-\lambda _{k})^{m_{k}}Q\,,$
where $Q$ has no zero. Then
    ${\begin{aligned}0&=\mu _{f}(f)\\&={\left((X-\lambda _{1})^{m_{1}}\cdots (X-\lambda _{k})^{m_{k}}Q\right)}(f)\\&=(f-\lambda _{1}\operatorname {Id} _{V})^{m_{1}}\cdots (f-\lambda _{k}\operatorname {Id} _{V})^{m_{k}}Q(f).\end{aligned}}$
We apply this mapping to $v$. Because of fact, the factors send the vector $v$ to $(\lambda -\lambda _{i})^{m_{i}}v$ or to $Q(\lambda )v$, respectively. Altogether, $v$ is sent to
    $(\lambda -\lambda _{1})^{m_{1}}\cdots (\lambda -\lambda _{k})^{m_{k}}Q(\lambda )v.$
As the composed mapping is the zero mapping and $Q(\lambda )\neq 0$, we must have $\lambda _{i}=\lambda$ for some $i$.
# Group theory/Order of an element/Definition

Let $G$ be a group and $g\in G$ an element. Then we call the smallest positive number $n$ with $g^{n}=e_{G}$ the order of $g$. For this, we write $[[:Template:Op:Ordergroupelement]]$. If all positive powers of $g$ are different from the neutral element, then we set $[[:Template:Op:Ordergroupelement]]=\infty$.
# Cyclotomic equation over C/Explicit description/Fact

Let $n\in \mathbb {N} _{+}$.
$X^{n}-1$
$\mathbb {C}$
    $e^{2\pi {\mathrm {i} }k/n}=\cos {\frac {2\pi k}{n}}+{\mathrm {i} }\sin {\frac {2\pi k}{n}},k=0,1,\ldots ,n-1.$
In $\mathbb {C} [X]$, we have the factorization
    $X^{n}-1=(X-1){\left(X-e^{2\pi {\mathrm {i} }/n}\right)}\cdots {\left(X-e^{2\pi {\mathrm {i} }(n-1)/n}\right)}\,.$
# Cyclotomic equation over C/Explicit description/Fact/Proof

The proof uses some basic facts about the complex exponential function. We have
    ${\left(e^{2\pi {\mathrm {i} }k/n}\right)}^{n}=e^{2\pi {\mathrm {i} }k}={\left(e^{2\pi {\mathrm {i} }}\right)}^{k}=1^{k}=1\,.$
Hence, the given complex numbers are indeed zeroes of the polynomial $X^{n}-1$. These zeroes are all different, because
    $e^{2\pi {\mathrm {i} }k/n}=e^{2\pi {\mathrm {i} }\ell /n}\,$
with $0\leq k\leq \ell \leq n-1$ implies, by considering the fraction, $e^{2\pi {\mathrm {i} }(\ell -k)/n}=1$ that
    $\ell -k=0\,$
holds. Therefore, there exist $n$ explicit zeroes and these are all the zeroes of the polynomial. The explicit description in coordinates follows from the Euler's formula.
# Permutation matrix/Definition

For a permutation $\pi$ on $\{1,\ldots ,n\}$, the $n\times n$-matrix
    $M_{\pi }={\left(a_{ij}\right)}\,,$
where
    $a_{\pi (j),j}=1\,$
and all other entries are $0$, is called a permutation matrix.
# Permutation matrix/Cycle/Characteristic polynomial/Fact/Proof

We may assume that the cycle has the form $1\mapsto 2\mapsto 3\mapsto \ldots \mapsto k\mapsto 1$. The corresponding permutation matrix $M_{\rho }$ looks with respect to $e_{k+1},\ldots ,e_{n}$ like the identity matrix and has, with respect to the first $k$ standard vectors, the form
    ${\begin{pmatrix}0&0&\ldots &0&1\\1&0&0&\ldots &0\\0&1&0&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&1&0\end{pmatrix}}.$
The determinant of $XE_{n}-M_{\rho }$ is $(X-1)^{n-k}$ multiplied with the determinant of
    ${\begin{pmatrix}X&0&\ldots &0&-1\\-1&X&0&\ldots &0\\0&-1&X&\ldots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\ldots &0&-1&X\end{pmatrix}}.$
The expansion with respect to the first row yields
    $X^{k}+(-1)^{k+1}(-1)(-1)^{k-1}=X^{k}-1\,.$
# Permutation matrix/Cycle/C/Eigenvalues/Fact

For a permutation matrix $M_{\rho }$ over $\mathbb {C}$ for a cycle $\rho \in S_{n}$ with $\rho :i_{1}\mapsto i_{2}\mapsto \ldots \mapsto i_{k}\mapsto i_{1}$
$k$
root of unity $\zeta$, the vectors
    $v_{\zeta }:=\zeta ^{k-1}e_{i_{1}}+\zeta ^{k-2}e_{i_{2}}+\cdots +\zeta e_{i_{k-1}}+e_{i_{k}}\,$
are eigenvectors of $M_{\rho }$ for the eigenvalue
$\zeta$
In particular, a permutation matrix of a cycle over $\mathbb {C}$ is diagonalizable.
# Permutation matrix/C/Diagonalizable/Fact

A permutation matrix
$\mathbb {C}$
# Cayley-Hamilton/Matrix version/Fact/Proof

We consider the matrix $XE_{n}-M$ as a matrix whose entries are in the field $K(X)$. The adjugate matrix
    $(XE_{n}-M)^{\operatorname {adj} }$
belongs also to $\operatorname {Mat} _{n}(K(X))$. The entries of the adjugate matrix are by definition the determinants of $(n-1)\times (n-1)$-submatrices of $XE_{n}-M$. In the entries of this matrix, the variable $X$ occurs at most in its first power, so that, in the entries of the adjugate matrix, the variable occurs at most in its $(n-1)$-th power. We write
    $(XE_{n}-M)^{\operatorname {adj} }=X^{n-1}A_{n-1}+X^{n-2}A_{n-2}+\cdots +XA_{1}+A_{0}\,$
with matrices
    $A_{i}\in \operatorname {Mat} _{n}(K)\,,$
that is, we write the entries as polynomials, and we collect all coefficients referring to $X^{i}$ into a matrix. Because of fact, we have
    ${\begin{aligned}\chi _{M}E_{n}&=(XE_{n}-M)\circ (XE_{n}-M)^{\operatorname {adj} }\\&=(XE_{n}-M)\circ (X^{n-1}A_{n-1}+X^{n-2}A_{n-2}+\cdots +XA_{1}+A_{0})\\&=X^{n}A_{n-1}+X^{n-1}(A_{n-2}-M\circ A_{n-1})+X^{n-2}(A_{n-3}-M\circ A_{n-2})+\cdots +X^{1}(A_{0}-M\circ A_{1})-M\circ A_{0}.\end{aligned}}$
We can write the matrix on the left according to the powers of $X$ and we get
    $\chi _{M}E_{n}=X^{n}E_{n}+X^{n-1}c_{n-1}E_{n}+X^{n-2}c_{n-2}E_{n}+\cdots +X^{1}c_{1}E_{n}+c_{0}E_{n}\,.$
Since these polynomials coincide, their coefficients coincide. That is, we have a system of equations
    ${\begin{matrix}E_{n}&=&A_{n-1}\\c_{n-1}E_{n}&=&A_{n-2}-M\circ A_{n-1}\\c_{n-2}E_{n}&=&A_{n-3}-M\circ A_{n-2}\\\vdots &\vdots &\vdots \\c_{1}E_{n}&=&A_{0}-M\circ A_{1}\\c_{0}E_{n}&=&-M\circ A_{0}\,.\end{matrix}}$
We multiply these equations from the left from top down with $M^{n},M^{n-1},M^{n-2},\ldots ,M^{1},E_{n}$, yielding the system of equations
    ${\begin{matrix}M^{n}&=&M^{n}\circ A_{n-1}\\c_{n-1}M^{n-1}&=&M^{n-1}\circ A_{n-2}-M^{n}\circ A_{n-1}\\c_{n-2}M^{n-2}&=&M^{n-2}\circ A_{n-3}-M^{n-1}\circ A_{n-2}\\\vdots &\vdots &\vdots \\c_{1}M^{1}&=&MA_{0}-M^{2}\circ A_{1}\\c_{0}E_{n}&=&-M\circ A_{0}\,.\end{matrix}}$
If we add the left-hand side of this system, then we just get $\chi _{M}\,(M)$. If we add the right-hand side, then we get $0$, because every partial summand $M^{i+1}\circ A_{i}$ occurs once positively and once negatively. Hence, we have $\chi _{M}\,(M)=0$.
# Polynomial ring/Field/1/Common divisor/Section
## Definition

Polynomial ring/K/Common divisor/Definition

## Definition

Polynomial ring/K/Greatest common divisor/Definition
A greatest common divisor is not uniquely determined, since with $G$ also $cG$ for some constant $c\neq 0$ is also a greatest common divisor. However, if we restrict to normed polynomials, then the greatest common divisor is unique.

## Definition

Polynomial ring/K/Relatively prime/Definition
# Polynomial ring/K/Common divisor/Definition

Let $P_{1},\ldots ,P_{n}\in K[X]$ denote polynomials over a field $K$. We say that a polynomial $T\in K[X]$ is a common divisor of the given polynomials if $T$ divides every $P_{i}$.
# Polynomial ring/K/Greatest common divisor/Definition

Let $P_{1},\ldots ,P_{n}\in K[X]$ denote polynomials over a field $K$. We say that a polynomial $G\in K[X]$ is a greatest common divisor of the given polynomials, if $G$ is a common divisor of the $P_{i}$, and if $G$ has among all common divisors of the $P_{i}$ maximal degree.
# Polynomial ring/K/Relatively prime/Definition

Polynomials $P_{1},\ldots ,P_{n}\in K[X]$ over a field $K$ are called coprime, if they have, with the exception of constants $c\neq 0$, no common divisor.
# Polynomial ring/Field/Lemma of Bezout/Section
## Theorem
### Proof

$\Box$

## Corollary
### Proof

$\Box$

## Remark

The method described in the previous remark is called Euclidean algorithm. It holds accordingly also for the integers.
# 3D Scan/Kinect Makerbot/Howard Community College/Spring2012/P1550jme/Javiereduturcios

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# 3D Scan/Kinect Makerbot/Howard Community College/Spring2012/P1550jme/Javiereduturcios/enes100

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