# Endomorphism/Polynomial/Inserting/Introduction/Section

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 $P\mapsto P(M)$.