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

$K[X]\longrightarrow \operatorname {End} _{}{\left(V\right)},P\longmapsto P(f),$

has the following properties.

1. For a constant polynomial ${}P=a_{0}$, we have

   ${}P(f)=a_{0}(f)=a_{0}f^{0}=a_{0}\operatorname {Id} _{V}\,.$

   In particular, the zero polynomial is sent to the zero mapping and the constant ${}1$-polynomial is sent to the identity.

2. We have

   ${}(P+Q)(f)=P(f)+Q(f)=Q(f)+P(f)\,$

   for all polynomials ${}P,Q\in K[X]$.

3. We have

   ${}(P\cdot Q)(f)=P(f)\circ Q(f)=Q(f)\circ P(f)\,$

   for all polynomials ${}P,Q\in K[X]$.

4. We have

   ${}(X^{n})(f)=f^{n}\,$

   for all ${}n\in \mathbb {N}$.

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:

• Does there exist some structure on the set of all polynomials
${}P\in K[X]$ with ${}P(f)=0$?
• Does there exist an especially simple polynomial
${}P_{0}\in K[X]$ with ${}P_{0}(f)=0$?
• How can we find it?
• Which properties of ${}f$ can we deduce from the factor decomposition of this polynomial ${}P_{0}$?

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 ${}P\mapsto P(M)$.