Endomorphism/Minimal polynomial/Introduction/Section

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

of ${}f$.

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

See exercise.

$\Box$

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 ${}M$.