# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 20

    The interpolation theorem
The following theorem is called theorem about polynomial interpolation and describes the interpolation of given function values by a polynomial. If just one function value at one point is given, then this determines a constant polynomial, two values at two points determine a linear polynomial (the graph is a line), three values at three points determine a quadratic polynomial, etc.

## Theorem

Let $K$ be a field, and let $n$ different elements $a_{1},\ldots ,a_{n}\in K$ and $n$ elements $b_{1},\ldots ,b_{n}\in K$ are given. Then there exist a unique
polynomial $P\in K[X]$ of degree $\leq n-1$, such that $P{\left(a_{i}\right)}=b_{i}$
$i$

### Proof

We prove the existence and consider first the situation where $b_{j}=0$ for all $j\neq i$ for some fixed $i$. Then
    $(X-a_{1})\cdots (X-a_{i-1})(X-a_{i+1})\cdots (X-a_{n})$
is a polynomial of degree $n-1$ which at the points $a_{1},\ldots ,a_{i-1},a_{i+1},\ldots ,a_{n}$ has value $0$. The polynomial
    ${\frac {b_{i}}{(a_{i}-a_{1})\cdots (a_{i}-a_{i-1})(a_{i}-a_{i+1})\cdots (a_{i}-a_{n})}}(X-a_{1})\cdots (X-a_{i-1})(X-a_{i+1})\cdots (X-a_{n})$
has at these points still a zero, but additionally at $a_{i}$ its value is $b_{i}$. We denote this polynomial by $P_{i}$. Then
    $P=P_{1}+P_{2}+\cdots +P_{n}\,$
is the polynomial looked for, because for the point $a_{i}$ we have
    $P_{j}(a_{i})=0\,$
for $j\neq i$ and $P_{i}(a_{i})=b_{i}$.
The uniqueness follows from fact.
$\Box$
A variant of the proof considers the mapping
    $K[X]\longrightarrow K^{n},P\longmapsto \left(P(a_{1}),\,\ldots ,\,P(a_{n})\right).$
This mapping is $K$-linear, since its components are linear due to remark. The interpolation theorem says that this mapping is surjective, this can be shown as in the proof. Moreover, it claims that this mapping, when restricted to the linear subspace of all polynomials of degree $\leq n-1$, is an isomorphism.

## Remark

If the data $a_{1},\ldots ,a_{n}$ and $b_{1},\ldots ,b_{n}$ are given, then one can find the interpolating polynomial $P$ of degree $\leq n-1$, which exists by fact, in the following way: We write
    $P=c_{0}+c_{1}X+c_{2}X^{2}+\cdots +c_{n-2}X^{n-2}+c_{n-1}X^{n-1}\,$
with unknown coefficients $c_{0},\ldots ,c_{n-1}$, and determine then these coefficients. Each interpolation point $(a_{i},b_{i})$ yields a linear equation
    $c_{0}+c_{1}a_{i}+c_{2}a_{i}^{2}+\cdots +c_{n-2}a_{i}^{n-2}+c_{n-1}a_{i}^{n-1}=b_{i}\,$
over $K$. The resulting system of linear equations has exactly one solution $(c_{0},\ldots ,c_{n-1})$, which gives the polynomial.
    Inserting an endomorphism
For a linear mapping
    $f\colon V\longrightarrow V$
on a $K$-vector space, we can consider the iterations $f^{n}$, that is, the $n$-fold composition of $f$ with itself. Moreover, we can add linear mappings and multiply them with scalars from the field. Therefore, expressions of the form
    $a_{n}f^{n}+a_{n-1}f^{n-1}+\cdots +a_{2}f^{2}+a_{1}f+a_{0}$
are linear mappings from $V$ to $V$. Here, we have to interpret
    $a_{0}=a_{0}f^{0}=a_{0}\operatorname {Id} _{V}\,.$
At first glance, it is not clear why studying such polynomial expressions in $f$ help in understanding $f$. The expression described can be understood in the sense that in the polynomial $a_{n}X^{n}+a_{n-1}X^{n-1}+\cdots +a_{2}X^{2}+a_{1}X+a_{0}$, we substitute the variable $X$ by the linear mapping $f$. This assignment fulfills the following structural properties.

## Lemma

Let $K$ be a field, $V$ a $K$-vector space, and
    $f\colon V\longrightarrow V$
a linear mapping. Then the mapping
    $K[X]\longrightarrow \operatorname {End} _{\left(V\right)},P\longmapsto P(f),$
* For a constant polynomial $P=a_{0}$, we have In particular, the zero polynomial is sent to the zero mapping and the constant $1$-polynomial is sent to the identity.
    $P(f)=a_{0}(f)=a_{0}f^{0}=a_{0}\operatorname {Id} _{V}\,.$
* We have for all polynomials $P,Q\in K[X]$.
    $(P+Q)(f)=P(f)+Q(f)=Q(f)+P(f)\,$
* We have for all polynomials $P,Q\in K[X]$.
    $(P\cdot Q)(f)=P(f)\circ Q(f)=Q(f)\circ P(f)\,$
* We have for all $n\in \mathbb {N}$.
    $(X^{n})(f)=f^{n}\,$

### Proof

(1) and (4) are inherent in the definition of the substitution homomorphism. From this, also (2) and (3) follows.
$\Box$
If $V$ is finite-dimensional, say of dimension $d$, then all powers $f^{k}$, $k\in \mathbb {N}$, are elements in the $d^{2}$-dimensional vector space
    $\operatorname {Hom} _{K}{\left(V,V\right)}=\operatorname {End} _{\left(V\right)}\,$
of all linear mappings from $V$ to $V$. Because the space of homomorphisms has also finite dimension, these powers must be linearly dependent. That is, there exists some $m\in \mathbb {N}$ and coefficients $a_{i}$, $0\leq i\leq m$, not all $0$, such that
    $a_{m}f^{m}+a_{m-1}f^{m-1}+\cdots +a_{2}f^{2}+a_{1}f+a_{0}=0\,$
holds (here, $m\leq d^{2}$ is immediately clear, we will see later that even $m\leq d$ holds). The corresponding polynomial $a_{m}X^{m}+a_{m-1}X^{m-1}+\cdots +a_{2}X^{2}+a_{1}X+a_{0}$ has the property that it is not the zero polynomial, and that, after replacing $X$ everywhere by $f$, the zero mapping on $V$ arises. We ask the following questions:

## Remark

Let $K$ be a field, $V$ a finite-dimensional $K$-vector space and
    $f\colon V\longrightarrow V$
a linear mapping. Let $v_{1},\ldots ,v_{n}$ be a basis of $V$, and let $M$ denote the corresponding matrix. Due to fact, we have a correspondence between compositions of linear mappings and matrix multiplication. In particular, $f^{n}$ corresponds with $M^{n}$. In the same way, the scalar multiplication and the addition on the space of endomorphisms $\operatorname {End} _{\left(V\right)}$ and on the space of matrices correspond to each other. Therefore, instead of the assignment $P\mapsto P(f)$, we can also work with the assignment $P\mapsto P(M)$.
    Ideals

## Definition

A subset ${\mathfrak {a}}$ of a commutative ring $R$ is called an ideal, if the following conditions are fulfilled:
* $0\in {\mathfrak {a}}$.
* For all $a,b\in {\mathfrak {a}}$, we have $a+b\in {\mathfrak {a}}$.
* For all $a\in {\mathfrak {a}}$ and $r\in R$, we have $ra\in {\mathfrak {a}}$.
The property $0\in {\mathfrak {a}}$ can be replaced by the condition that ${\mathfrak {a}}$ is not empty. An ideal is a subgroup of the additive group of $R$, which, moreover, is also closed under scalar multiplication.

## Definition

For a family of element $a_{1},a_{2},\ldots ,a_{n}\in R$ in a commutative ring $R$, we denote by $(a_{1},a_{2},\ldots ,a_{n})$ the ideal generated by these elements. It consists of all linear combinations
    $r_{1}a_{1}+r_{2}a_{2}+\cdots +r_{n}a_{n},$
where
$r_{1},r_{2},\ldots ,r_{n}\in R$

## Definition

An ideal ${\mathfrak {a}}$ in a commutative ring $R$ of the form
    ${\mathfrak {a}}=(a)=Ra=\{ra:\,r\in R\}\,$
The zero element forms, in every ring, the so-called zero ideal. We write this simply as $0=(0)=\{0\}$. write. The $1$ and, moreover, every unit, generates as an ideal the total ring. A unit in a commutative ring $R$ is an invertible element, that is, an element $x\in R$ such that there exists an $y\in R$ with $xy=1$. A commutative ring is a field if and only if all elements with the exception of $0$ are units.

## Definition

The unit ideal in a commutative ring
$R$
In a field, there exist exactly two ideals.

## Lemma

Let $R$ be a commutative ring. Then the following statements are equivalent.
* $R$ is a field.
* There exist exactly two ideals in $R$.

### Proof

If $R$ is a field, then there exists the zero ideal and the unit ideal, and these are different ideals. Let $I$ be an ideal in $R$ different from $0$. Then $I$ contains some element $x\neq 0$, which is a unit. Therefore, $1=xx^{-1}\in I$ and thus $I=R$.
Suppose now that $R$ is a commutative ring with exactly two ideals. Then $R$ is not the zero ring. Let now $x$ be an element in $R$ different from $0$. The principal ideal generated by $x$, that is $Rx$, is $\neq 0$, and therefore it must be the other ideal, which is the unit ideal. In particular, this means $1\in Rx$. Hence, $1=xr$ for some $r\in R$, so that $x$ is a unit.
$\Box$
    Ideals in $K[X]$

## Theorem

In a polynomial ring over a field, every ideal is a

### Proof

Let $I$ denote an ideal in $K[X]$ different from $0$. We consider the non-empty set of natural numbers
    ${\left\{\operatorname {deg} \,(P)\mid P\in I,\,P\neq 0\right\}}.$
This set has a minimum $m\in \mathbb {N}$. This number arises from an element $F\in I$, $F\neq 0$, $m=\operatorname {deg} \,(F)$. We claim that $I=(F)$.
The inclusion $\supseteq$ is clear. To prove the other inclusion $\subseteq$, let $P\in I$ be given. Due to fact, we have
    $P=FQ+R{\text{ and with }}\operatorname {deg} \,(R)<\operatorname {deg} \,(F){\text{ or }}R=0.$
Because of $R\in I$ and the minimality of $\operatorname {deg} \,(F)$, the first case can not happen. Therefore, $R=0$, and this means that $P$ is a multiple of $F$.
$\Box$
    The minimal polynomial

## Definition

Let $V$ be a finite-dimensional $K$-vector space and
    $f\colon V\longrightarrow V$
a linear mapping. Then the uniquely determined normed polynomial $\mu _{f}\in K[X]$ of minimal degree fulfilling
    $\mu _{f}(f)=0\,$
is called the minimal polynom
$f$

## Corollary

Let $V$ be a finite-dimensional vector space over a field $K$, and let
    $f\colon V\longrightarrow V$
denote a linear mapping. Then the set
    ${\left\{P\in K[X]\mid P(f)=0\right\}}$
is a principal ideal in the polynomial ring $K[X]$, which is generated by the minimal polynomial
$\mu _{f}$

### Proof

$\Box$

## Example

For the identity $\operatorname {Id} _{V}$ on a $K$-vector space $V$, the minimal polynomial is just $X-1$. This polynomial is sent under the evaluation homomorphism to
    $\operatorname {Id} _{V}-\operatorname {Id} _{V}=0\,.$
A constant polynomial $a_{0}$ is sent to $a_{0}\operatorname {Id}$, which is not, with the exception of $a_{0}=0$ or $V=0$, the zero mapping.
For a homothety, that is, a mapping of the form $\lambda \operatorname {Id} _{V}$, the minimal polynomial is $X-\lambda$, under the condition $\lambda \neq 0$ and $V\neq 0$. For the zero mapping on $V\neq 0$, the minimal polynomial is $X$, in case $V=0$, it is the constant polynomial $1$.

## Example

For a diagonal matrix
    $M={\begin{pmatrix}d_{1}&0&\cdots &0\\0&d_{2}&\ddots &\vdots \\\vdots &\ddots &\ddots &0\\0&\cdots &0&d_{n}\end{pmatrix}}\,$
with different entries $d_{i}$, the minimal polynomial is
    $P=(X-d_{1})(X-d_{2})\cdots (X-d_{n})\,.$
This polynomial is sent under the substitution to
    $(M-d_{1}E_{n})\circ (M-d_{2}E_{n})\circ (M-d_{n}E_{n}).$
We apply this to a standard vector $e_{i}$. Then factor $(M-d_{j}E_{n})$ sends $e_{i}$to $(d_{i}-d_{j})e_{i}$. Therefore, the $i$-th factor ensures that $e_{i}$ is annihilated. Since a basis is mapped under $P(M)$ to $0$, it must be the zero mapping.
Assume now that $P$ is not the minimal polynomial $\mu$. Then there exists, due to fact, a polynomial $Q$ with
    $P=Q\mu \,,$
and, because of fact, $\mu$ is a partial product of the linear factors of $P$. But as soon as one factor of $P$ is removed, say we remove $X-d_{i}$, then $e_{i}$ is not annihilated by the corresponding mapping.

## Example

For the matrix
    $M={\begin{pmatrix}0&1\\0&0\end{pmatrix}}\,,$
$X^{2}$ is the minimal polynomial. This polynomial becomes, after the substitution, the zero mapping, because of
    $M^{2}={\begin{pmatrix}0&0\\0&0\end{pmatrix}}\,.$
The factors of $X^{2}$ of smaller degree are the constant polynomials $\neq 0$ and $a_{1}X$ with $a_{1}\neq 0$, but these polynomials do not annihilate $M$.