Endomorphism/Nilpotent/Introduction/Section

Definition

Let ${}K$ be a field, and let ${}V$ be a ${}K$-vector space. A linear mapping

$\varphi \colon V\longrightarrow V$

is called nilpotent, if there exists a natural number ${}n$ such that the ${}n$-th composition fulfills

${}\varphi ^{n}=0\,.$

Definition

A square matrix ${}M$ is called nilpotent, if there exists a natural number ${}n\in \mathbb {N}$ such that the ${}n$-th matrix product fulfills

${}M^{n}=\underbrace {M\circ \cdots \circ M} _{n{\text{-mal}}}=0\,$

Example

Let ${}M$ be an upper triangular matrix with the property that all diagonal entries are ${}0$. Thus, ${}M$ has the form

${\begin{pmatrix}0&*&\cdots &\cdots &*\\0&0&*&\cdots &*\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&0&*\\0&\cdots &\cdots &0&0\end{pmatrix}}.$

Then ${}M$ is nilpotent, with every power the ${}0$-diagonal is moved one step up and to the right. If, for example, the product of the ${}i$-th row and the ${}j$-th column with

${}i\geq j-1\,$

is computed, then there is always a ${}0$ in the partial products and, altogether, the result is ${}0$.

Example

A special case of example is the matrix

${\begin{pmatrix}0&1&0&\cdots &\cdots &0\\0&0&1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&0&1&0\\0&\cdots &\cdots &0&0&1\\0&\cdots &\cdots &\cdots &0&0\end{pmatrix}}.$

An important observation is that under this mapping, ${}e_{n}$ is sent to ${}e_{n-1}$, ${}e_{n-1}$ is sent to ${}e_{n-2}$, and , finally, ${}e_{2}$ is sent to ${}e_{1}$, which is sent to ${}0$. The ${}(n-1)$-th power of the matrix sends ${}e_{n}$ to ${}e_{1}$ and is not the zero matrix, but the ${}n$-th power of the matrix is the zero matrix.

Example

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a linear mapping. For an eigenvalue ${}\lambda \in K$, the generalized eigenspace ${}H=\operatorname {GeEig} _{\lambda }(\varphi )$ has the property that the restriction of ${}\varphi -\lambda \operatorname {Id} _{V}$ to ${}H$ is nilpotent.

Lemma

Let ${}V$ denote a finite-dimensional vector space over a field ${}K$. Let

$\varphi \colon V\longrightarrow V$

be a

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is nilpotent.
2. For every vector ${}v\in V$, there exists an ${}k\in \mathbb {N}$ such that

   ${}\varphi ^{k}(v)=0\,.$

3. There exists a basis ${}v_{1},\ldots ,v_{n}$ of ${}V$ and a ${}k\in \mathbb {N}$ such that

   ${}\varphi ^{k}(v_{i})=0\,$

   for ${}i=1,\ldots ,n$.

4. There exists a generating system ${}v_{1},\ldots ,v_{m}$ of ${}V$ and a ${}k\in \mathbb {N}$ such that

   ${}\varphi ^{k}(v_{i})=0\,$

   for ${}i=1,\ldots ,n$.

Proof

From (1) to (2) is clear. From (2) to (3). Let ${}v_{1},\ldots ,v_{n}$ be a basis (or a finite generating system), and let ${}k_{j}\in \mathbb {N}$ be such that

${}\varphi ^{k_{j}}(v_{j})=0\,.$

Then

${}k:={\max {\left(k_{j},j=1,\ldots ,m\right)}}\,$

fulfills the property for every generator. From (3) to (4) is clear. From (4) to (1). For ${}v\in V$, we have

${}v=\sum _{i=1}^{m}a_{i}v_{i}\,.$

Due to the linearity of ${}\varphi ^{k}$, we have

${}\varphi ^{k}(v)=\varphi ^{k}{\left(\sum _{i=1}^{m}a_{i}v_{i}\right)}=\sum _{i=1}^{m}a_{i}\varphi ^{k}(v_{i})=0\,,$

therefore,

${}\varphi ^{k}=0\,.$

$\Box$

Lemma

Let ${}K$ denote a field, and let ${}V$ denote a ${}K$-vector space of finite dimension. Let

$\varphi \colon V\longrightarrow V$

be a

linear mapping. Then the following statements are equivalent.

1. ${}\varphi$ is nilpotent
2. The minimal polynomial of ${}\varphi$ is a power of ${}X$.
3. The characteristic polynomial of ${}\varphi$ is a power of ${}X$.

Proof

The equivalence of (1) and (2) follows immediately from the definition, the equivalence of (2) and (3) follows from fact.

$\Box$

Corollary

Let ${}K$ be a field and let ${}V$ denote a finite-dimensional ${}K$-vector space. Let

$\varphi \colon V\longrightarrow V$

be a nilpotent linear mapping. Then ${}\varphi$ is

trigonalizable. There exists a basis such that ${}\varphi$ is described, with respect to this basis, by an upper triangular matrix, in which all diagonal entries are ${}0$.

Proof

This follows directly from fact and fact.

$\Box$