# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 19

In the lectures to come we will try to understand a
$d\times d$-matrix $M$ (an endomorphism) by looking at expressions of the form
    $a_{n}M^{n}+a_{n-1}M^{n-1}+\cdots +a_{2}M^{2}+a_{1}M^{1}+a_{0}M^{0},$
where $M^{i}$ means the $i$-th matrix product of the matrix with itself, $M^{0}$being the unit matrix $E_{d}$. Such expressions arise by plugging matrices into polynomials. In this lecture, we introduce polynomials and the polynomial ring.
    The polynomial ring over a field

## Definition

The polynomial ring over a field $K$ consists of all polynomials
    $P=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,,$
where $a_{i}\in K$, $n\in \mathbb {N}$. It is endowed with componentwise addition and a multiplication, which arises by distributive continuation of the rule
    $X^{n}\cdot X^{m}:=X^{n+m}\,.$
A polynomial
    $P=\sum _{i=0}^{n}a_{i}X^{i}=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,$
is, formally seen, nothing but the tuple $(a_{0},a_{1},\ldots ,a_{n})$, these numbers are called the coefficient of the polynomial. The polynomials are equal if and only if all their coefficients coincide. The letter $X$ is called the variable of the polynomial ring. In this context, the field $K$ is called the base field of the polynomial ring. Due to the componentwise definition of the addition, we have immediately a commutative group, with the zero polynomial (where all coefficients are $0$) as neutral element. The polynomials with $a_{i}=0$ for all $i\geq 1$ are called constant polynomials, they are simply written as $a_{0}$.
The way a polynomial is written suggests how the multiplication shall work, the product $X^{n}\cdot X^{m}$ is given by the addition of the exponents, thus $X^{n}\cdot X^{m}:=X^{n+m}$. For arbitrary polynomials, the multiplication arises from this simple multiplication rule by distributive continuation according to the law to multiply "everything with everything“. Explicitly, the multiplication is given by the following rule gegeben:
    ${\left(\sum _{i=0}^{n}a_{i}X^{i}\right)}\cdot {\left(\sum _{j=0}^{m}a_{j}X^{j}\right)}=\sum _{k=0}^{n+m}c_{k}X^{k}{\text{ where }}c_{k}=\sum _{r=0}^{k}a_{r}b_{k-r}.$
The multiplication is associative, commutative, distributive, and the constant polynomial $1$ is its neutral element, see exercise. Altogether, we have a commutative ring.

## Definition

The degree of a nonzero polynomial
    $P=a_{0}+a_{1}X+a_{2}X^{2}+\cdots +a_{n}X^{n}\,$
with $a_{n}\neq 0$
$n$
We do not define a degree for the zero polynomial. The coefficient $a_{n}$, where $n$ is the degree of the polynomial, is called the leading coefficient of the polynomial. The term $a_{n}X^{n}$ is called leading term. A polynomial with leading coefficient $1$ is called normed.
We can plug in (or insert or evaluate at) an element $a\in K$, into a polynomial $P\in K[X]$, by replacing the variable $X$ everywhere by $a$. This gives a mapping
    $K\longrightarrow K,a\longmapsto P(a),$
which we call the polynomial function defined by the polynomial. In general, this mapping is not linear, only the polynomials of the form $P=a_{1}X$ are linear.
    Die Division with remainder

## Definition

Let $K$ be a field. We say that a polynomial $T\in K[X]$ divides a polynomial $P\in K[X]$, if there exists a polynomial $Q\in K[X]$ such that
    $P=TQ\,.$
If $P$ can be divided by $T$, we also say that $P$ is a multiple of $T$. In $K[X]$ it is not possible to divide any element by any element $\neq 0$. This is different from a field, but similar no the integers $\mathbb {Z}$. However, there is an important substitute, the Euclidean division.

## Theorem

Let $K$ be a field and let $K[X]$ be the polynomial ring over $K$. Let $P,T\in K[X]$ be polynomials with $T\neq 0$. Then there exist unique polynomials $Q,R\in K[X]$ such that
    $P=TQ+R{\text{ and with }}\operatorname {deg} \,(R)<\operatorname {deg} \,(T){\text{ or }}R=0.$

### Proof

We prove the statement about the existence by induction over the degree of $P$. If the degree of $T$ is larger than the degree of $P$, then $Q=0$ and $R=P$ is a solution.
Suppose that $\operatorname {deg} \,(P)=0$. By the remark just made also $\operatorname {deg} \,(T)=0$ holds, so $T$ is a constant polynomial, and therefore (since $T\neq 0$ and $K$ is a field) $Q=P/T$ and $R=0$ is a solution.
So suppose now that $\operatorname {deg} \,(P)=n$ and that the statement for smaller degrees is already proven. We write $P=a_{n}X^{n}+\cdots +a_{1}X+a_{0}$ and $T=b_{k}X^{k}+\cdots +b_{1}X+b_{0}$ with $a_{n},b_{k}\neq 0,\,k\leq n$. Then setting $H={\frac {a_{n}}{b_{k}}}X^{n-k}$ we have the relation
    ${\begin{aligned}P'&:=P-TH\\&=0X^{n}+{\left(a_{n-1}-{\frac {a_{n}}{b_{k}}}b_{k-1}\right)}X^{n-1}+\cdots +{\left(a_{n-k}-{\frac {a_{n}}{b_{k}}}b_{0}\right)}X^{n-k}+a_{n-k-1}X^{n-k-1}+\cdots +a_{0}.\end{aligned}}$
The degree of this polynomial $P'$ is smaller than $n$ and we can apply the induction hypothesis to it. That means there exist $Q'$ and $R'$ such that
    $P'=TQ'+R'{\text{ and with }}\operatorname {deg} \,(R')<\operatorname {deg} \,(T){\text{ or }}R'=0.$
From this we get altogether
    $P=P'+TH=TQ'+TH+R'=T(Q'+H)+R'\,,$
so that $Q=Q'+H$ and $R=R'$ is a solution.
To prove uniqueness, let $P=TQ+R=TQ'+R'$, both fulfilling the stated conditions. Then $T(Q-Q')=R'-R$. Since the degree of the difference $R'-R$ is smaller than $\operatorname {deg} \,(T)$, this implies $R=R'$ and so $Q=Q'$.
$\Box$
The polynomial $T$ divides $P$ if and only if the remainder in the Euclidean division is $0$. The proof of this theorem is constructive, that is, it describes a method how to perform the Euclidean division effectively. For this, it is necessary to be able to perform the operations in the base field. We give two examples, one over the rational numbers and one over the complex numbers.

## Example

We want to apply the Euclidean division (over $\mathbb {Q}$)
    $P=6X^{3}+X+1{\text{ divided by }}T=3X^{2}+2X-4.$
So we want to divide a polynomial of degree $3$ by a polynomial of degree $2$, hence the quotient and also the remainder have (at most) degree $1$. For the first step, we ask with which term we have to multiply $T$ to achieve that the product and $P$ have the same leading term. This is $2X$. The product is
    $2X{\left(3X^{2}+2X-4\right)}=6X^{3}+4X^{2}-8X\,.$
The difference between $P$ and this product is
    $6X^{3}+X+1-{\left(6X^{3}+4X^{2}-8X\right)}=-4X^{2}+9X+1\,.$
We continue the division by $T$ with this polynomial, which we call $P'$. In order to get coincidence with the leading coefficient we have to multiply $T$ with ${\frac {-4}{3}}$. This yields
    $-{\frac {4}{3}}T=-{\frac {4}{3}}{\left(3X^{2}+2X-4\right)}=-4X^{2}-{\frac {8}{3}}X+{\frac {16}{3}}\,.$
The difference between this and $P'$ is therefore
    $-4X^{2}+9X+1-{\left(-4X^{2}-{\frac {8}{3}}X+{\frac {16}{3}}\right)}={\frac {35}{3}}X-{\frac {13}{3}}\,.$
This is the remainder and altogether we get
    $6X^{3}+X+1={\left(3X^{2}+2X-4\right)}{\left(2X-{\frac {4}{3}}\right)}+{\frac {35}{3}}X-{\frac {13}{3}}\,.$

## Example

We perform the Euclidean division
    $P=(4+3{\mathrm {i} })X^{3}+X^{2}+5{\mathrm {i} }{\text{ divided by }}T=(1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }.$
The inverse of $1+{\mathrm {i} }$ is ${\frac {1}{2}}-{\frac {1}{2}}{\mathrm {i} }$, and therefore we have
    ${\begin{aligned}(4+3{\mathrm {i} })(1+{\mathrm {i} })^{-1}&=(4+3{\mathrm {i} }){\left({\frac {1}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}\\&=2+{\frac {3}{2}}-2{\mathrm {i} }+{\frac {3}{2}}{\mathrm {i} }\\&={\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }.\end{aligned}}$
Hence, $Q$ starts with ${\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X$, and we have
    ${\begin{aligned}((1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }){\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X&=(4+3{\mathrm {i} })X^{3}+{\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-{\frac {19}{2}}+{\frac {17}{2}}{\mathrm {i} }\right)}X\\\end{aligned}}$
We have to subtract this term from $P$ and we obtain
    ${\begin{aligned}P-{\left((4+3{\mathrm {i} })X^{3}+{\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-{\frac {19}{2}}+{\frac {17}{2}}{\mathrm {i} }\right)}X\right)}&={\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left({\frac {19}{2}}-{\frac {17}{2}}{\mathrm {i} }\right)}X+5{\mathrm {i} }\\\,\end{aligned}}$
We apply the same procedure to this polynomial (which we call $P'$). We compute
    ${\begin{aligned}{\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}{\left({\frac {1}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}&=-1+{\frac {3}{2}}{\mathrm {i} }\\\end{aligned}}$
Therefore, the constant term of $Q$ equals $-1+{\frac {3}{2}}{\mathrm {i} }$, and we obtain
    ${\left((1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }\right)}{\left(-1+{\frac {3}{2}}{\mathrm {i} }\right)}={\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-1+{\frac {3}{2}}{\mathrm {i} }\right)}X-{\frac {13}{2}}{\mathrm {i} }\,.$
We subtract this from $P'$ and get
    $P'-{\left({\left(-{\frac {5}{2}}+{\frac {1}{2}}{\mathrm {i} }\right)}X^{2}+{\left(-1+{\frac {3}{2}}{\mathrm {i} }\right)}X-{\frac {13}{2}}{\mathrm {i} }\right)}={\left({\frac {21}{2}}-10{\mathrm {i} }\right)}X+{\frac {23}{2}}{\mathrm {i} }\,.$
This term is the remainder $R$, the Euclidean division altogether is
    ${\begin{aligned}(4+3{\mathrm {i} })X^{3}+X^{2}+5{\mathrm {i} }&=((1+{\mathrm {i} })X^{2}+X-3+2{\mathrm {i} }){\left({\left({\frac {7}{2}}-{\frac {1}{2}}{\mathrm {i} }\right)}X-1+{\frac {3}{2}}{\mathrm {i} }\right)}+{\left({\frac {21}{2}}-10{\mathrm {i} }\right)}X+{\frac {23}{2}}{\mathrm {i} }\\\,\end{aligned}}$
    Zeroes
A zero of a polynomial $P$ is an element $a\in K$ such that $P(a)=0$. A polynomial does not necessarily have zeroes, and this depends also on the base field. The polynomial $X^{2}+1$ has no real zero, but it has the complex zeroes ${\mathrm {i} }$ and $-{\mathrm {i} }$. As an element in $\mathbb {R} [X]$, the polynomial $X^{2}+1$ can not be written as a product of simpler polynomials. However, in $\mathbb {C} [X]$, it has the factor decomposition
    $X^{2}+1=(X-{\mathrm {i} })(X+{\mathrm {i} })\,.$

## Remark

Let $K$ be a field, let $K[X]$ be the polynomial ring over $K$ and $a\in K$. Then the evaluation mapping
    $K[X]\longrightarrow K,P\longmapsto P(a),$
is $K$-linear. Moreover, we have
    $(PQ)(a)=P(a)Q(a)\,,$
see exercise.

## Lemma

Let $K$ be a field and let $K[X]$ be the polynomial ring over $K$. Let $P\in K[X]$ be a polynomial and $a\in K$. Then $a$ is a zero
$P$
$P$
$X-a$

### Proof

If $P$ is a multiple of $X-a$, then we can write
    $P=(X-a)Q\,$
with another polynomial $Q$. Inserting $a$ yields
    $P(a)=(a-a)Q(a)=0\,.$
In general, there exists, due to fact, a representation
    $P=(X-a)Q+R\,,$
where either $R=0$ or the degree of $R$ is $0$, so in both cases $R$ is a constant. Inserting $a$ yields
    $P(a)=R\,.$
So if $P(a)=0$ holds, then the remainder must be $R=0$, and this means $P=(X-a)Q$.
$\Box$

## Corollary

Let $K$ be a field and let $K[X]$ be the polynomial ring over $K$. Let $P\in K[X]$ be a polynomial ($\neq 0$) of degree
$d$
$P$
$d$

### Proof

We prove the statement by induction over $d$. For $d=0,1$ the statement holds. So suppose that $d\geq 2$ and that the statement is already proven for smaller degrees. Let $a$ be a zero of $P$ (if $P$ does not have a zero at all, we are done anyway). Hence, $P=Q(X-a)$ by fact and the degree of $Q$ is $d-1$, so we can apply to $Q$ the induction hypothesis. The polynomial $Q$ has at most $d-1$ zeroes. For $b\in K$ we have $P(b)=Q(b)(b-a)$. This can be zero, due to fact, only if one factor is $0$, so the zeroes of $P$ are $a$ or a zero of $Q$. Hence, there are at most $d$ zeroes of $P$.
$\Box$
    The Fundamental theorem of algebra
The following Fundamental theorem of algebra holds, which we state without a proof.

## Theorem

Every nonconstant polynomial $P\in \mathbb {C} [X]$ over the complex numbers has a
The Fundamental theorem of algebra implies that every polynomial $P\in \mathbb {C} [X]$ different from $0$ has a factorization into linear factors, that is,
    $P=c(X-z_{1})(X-z_{2})\cdots (X-z_{n})\,,$
where, up to the ordering, the complex numbers $c,z_{1},\ldots ,z_{n}$ are uniquely determined (and repetitions may happen).
    Rational functionen
The polynomial ring $K[X]$ is a commutative ring, but not a field. However, we can construct a field which contains the polynomial ring with the help of the so-called formal-rational functions, in a similar way as we can construct the rational numbers $\mathbb {Q}$ from the integers $\mathbb {Z}$. For this, we define
    $K(X):={\left\{{\frac {P}{Q}}\mid P,Q\in K[X],\,Q\neq 0\right\}}\,,$
where we identify, like in $\mathbb {Q}$, two fractions ${\frac {P}{Q}}$ and ${\frac {P'}{Q'}}$, whenever
    $PQ'=P'Q\,$
holds. In this way, the field of rational functions (over $K$) arises.
The formal expression $P/Q$ can be considered as a function in the following way.

## Definition

Let $K$ be a field. For polynomials $P,Q\in K[X]$, $Q\neq 0$, the function
    $D\longrightarrow K,z\longmapsto {\frac {P(z)}{Q(z)}},$
where $D$ is the complement of the zeroes
$Q$
Next to the polynomial functions, the simplest functions are the rational functions.
    Footnotes
* ↑ Here, like for the addition of polynomials of different degrees, the coefficients for $r>n$ or $k-r>m$ are $0$.