# Endomorphism/Nilpotent/Jordan decomposition/Section

For a nilpotent endomorphism $\varphi$ on a vector space $V$, we have
    $V=\operatorname {GeEig} _{0}(\varphi )\,,$
hence, there is just one
generalized eigenspace, and this is the total space. We will show that we can improve the describing matrix even further (not just having triangular form).

## Example

A matrix of the form
    $M={\begin{pmatrix}0&a\\0&0\end{pmatrix}}\,$
with $a\neq 0$ has, with respect the basis $ae_{1}$ and $e_{2}$, the form
    ${\begin{pmatrix}0&1\\0&0\end{pmatrix}}.$

## Lemma

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. Let
    $\varphi ^{s}=0\,$
and suppose that $s$ is minimal with this property. Then, between the linear subspaces
    $V_{i}:=\operatorname {kern} \varphi ^{i}\,$
the relation
    $\varphi ^{-1}(V_{i})=V_{i+1}\,$
holds, and the inclusions
    $V_{i}\subset V_{i+1}\,$
are strict for
$i<s$

### Proof

Let $v\in V$. Then, the containment $v\in V_{i+1}=\operatorname {kern} \varphi ^{i+1}$ is equivalent with $\varphi (v)\in V_{i}=\operatorname {kern} \varphi ^{i}$. This gives the first claim. For the second claim, assume that
    $V_{i}=V_{i+1}\,$
holds for some $i<s$. By applying $\varphi ^{-1}$, we get
    $V_{i+1}=\varphi ^{-1}(V_{i})=\varphi ^{-1}(V_{i+1})=V_{i+2}\,.$
In this way, we obtain
    $V_{i}=V_{i+1}=V_{i+2}=\ldots =V_{s}=V\,,$
contradicting the minimality of $s$.
$\Box$

## Lemma

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 there exists a basis $v_{1},\ldots ,v_{n}$ of $V$ with
    $\varphi (v_{j})=v_{j-1}\,$
or
    $\varphi (v_{j})=0\,.$

### Proof

Let $\varphi ^{s}=0$ and suppose that $s$ is minimal with this property. We consider the linear subspaces
    $V_{i}:=\operatorname {kern} \varphi ^{i}\,.$
Let $U_{s}$ be a direct complement for $V_{s-1}$, therefore,
    $V=V_{s-1}\oplus U_{s}\,.$
Because of fact, we have
    $\varphi ^{-1}(V_{s-2})=V_{s-1}\,$
and
    $\varphi (U_{s})\cap V_{s-2}=0\,.$
Therefore, there exists a linear subspace $U_{s-1}$ of $V_{s-1}$ with
    $V_{s-1}=V_{s-2}\oplus U_{s-1}\,$
and with
    $\varphi (U_{s})\subseteq U_{s-1}\,.$
In this way, we obtain linear subspaces $U_{i}\subseteq V_{i}$ such that
    $V_{i}=V_{i-1}\oplus U_{i}\,$
and
    $\varphi (U_{i})\subseteq U_{i-1}\,.$
Morover,
    $V=U_{1}\oplus \cdots \oplus U_{s}\,,$
since we refine the preceding direct sum decomposition in every step. Also, $\varphi$, restricted to $U_{i}$ with $i\geq 2$ is injective. For $v\in U_{i}\cap \operatorname {kern} \varphi =U_{i}\cap V_{1}\subseteq U_{i}\cap V_{i-1}$, it follows
    $v=0\,$
by the directness of the composition. We construct now a basis with the claimed properties. For this, we choose a basis ${\mathfrak {u}}_{s}$ of $U_{s}$. We can extend the (linearly independent) image $\varphi ({\mathfrak {u}}_{s})$ to get a basis ${\mathfrak {u}}_{s-1}$ of $U_{s-1}$, and so forth. The union of these bases is then a basis of $V$. The basis element of ${\mathfrak {u}}_{i}$ for $i\geq 2$ are sent by construction to other basis elements, and the basis elements of ${\mathfrak {u}}_{1}$ are sent to $0$. To get an ordering, we choose a basis element from ${\mathfrak {u}}_{s}$, together with all its successive images, then we choose another basis element of ${\mathfrak {u}}_{s}$, together with all its successive images, until the ${\mathfrak {u}}_{s}$ is exhausted. Then we work with ${\mathfrak {u}}_{s-1}$ in the same way. In the last step, we swap the ordering of the basis elements just constructed.
$\Box$

## 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 there exists a basis of $V$ such that describing matrix, with respect to this basis, has the form
    ${\begin{pmatrix}0&c_{1}&0&\cdots &\cdots &0\\0&0&c_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&0&c_{n-2}&0\\0&\cdots &\cdots &0&0&c_{n-1}\\0&\cdots &\cdots &\cdots &0&0\end{pmatrix}},$
$c_{i}$
$0$
$1$
$\varphi$

### Proof

This follows directly from fact.
$\Box$
For a nilpotent mapping on a two-dimensional vector space $V$, we either have the zero mapping, or a nilpotent mapping with an one-dimensional kernel. In this case, we obtain man for every element $v\in V\setminus \operatorname {kern} \varphi$ a basis $\varphi (v),v$ (in this ordering), such that the describing matrix has the form ${\begin{pmatrix}0&1\\0&0\end{pmatrix}}$. When the dimension is larger,we get more and more complex possibilities. We discuss some typical examples in dimension three.

## Example

We want to apply fact to
    $M={\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}\,.$
We have
    $M^{2}={\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}\cdot {\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}={\begin{pmatrix}0&0&10\\0&0&0\\0&0&0\end{pmatrix}}\,$
and
    $M^{3}=0\,.$
Therefore,
    $V_{1}=\operatorname {kern} M=\langle e_{1}\rangle ,\,V_{2}=\operatorname {kern} M^{2}=\langle e_{1},e_{2}\rangle {\text{ and }}V_{3}=K^{3}.$
We have
    $V_{3}=V_{2}\oplus \langle e_{3}\rangle \,,$
so that wir can choose
    $U_{3}=\langle e_{3}\rangle \,.$
We have
    $Me_{3}={\begin{pmatrix}3\\5\\0\end{pmatrix}}\in V_{2}\,.$
Hence,
    $V_{2}=V_{1}\oplus U_{2}\,$
with
    $U_{2}=\langle {\begin{pmatrix}3\\5\\0\end{pmatrix}}\rangle \,.$
Finally,
    $M^{2}e_{3}=M{\begin{pmatrix}3\\5\\0\end{pmatrix}}={\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}{\begin{pmatrix}3\\5\\0\end{pmatrix}}={\begin{pmatrix}10\\0\\0\end{pmatrix}}\,.$
Therefore,
    ${\begin{pmatrix}10\\0\\0\end{pmatrix}},\,{\begin{pmatrix}3\\5\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\0\\1\end{pmatrix}}$
is a basis with the intended properties.
The inverse matrix to
    $B={\begin{pmatrix}10&3&0\\0&5&0\\0&0&1\end{pmatrix}}\,$
is
    ${\begin{pmatrix}{\frac {1}{10}}&-{\frac {3}{50}}&0\\0&{\frac {1}{5}}&0\\0&0&1\end{pmatrix}},$
therefore,
    ${\begin{aligned}B^{-1}MB&={\begin{pmatrix}{\frac {1}{10}}&-{\frac {3}{50}}&0\\0&{\frac {1}{5}}&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0&2&3\\0&0&5\\0&0&0\end{pmatrix}}{\begin{pmatrix}10&3&0\\0&5&0\\0&0&1\end{pmatrix}}\\&={\begin{pmatrix}{\frac {1}{10}}&-{\frac {3}{50}}&0\\0&{\frac {1}{5}}&0\\0&0&1\end{pmatrix}}{\begin{pmatrix}0&10&3\\0&0&5\\0&0&0\end{pmatrix}}\\&={\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}}.\end{aligned}}$

## Example

We want to apply fact to
    $M={\begin{pmatrix}0&0&3\\0&0&7\\0&0&0\end{pmatrix}}\,.$
We have
    $M^{2}=0\,.$
Therefore,
    $V_{1}=\operatorname {kern} M=\langle e_{1},e_{2}\rangle {\text{ and }}V_{2}=K^{3}.$
We have
    $V_{2}=V_{1}\oplus \langle e_{3}\rangle \,,$
so that we can choose
    $U_{2}=\langle e_{3}\rangle \,.$
We have
    $Me_{3}={\begin{pmatrix}3\\7\\0\end{pmatrix}}\in V_{1}\,.$
Therefore,
    $V_{1}=\langle {\begin{pmatrix}3\\7\\0\end{pmatrix}},e_{1}\rangle =U_{1}\,.$
Hence,
    ${\begin{pmatrix}1\\0\\0\end{pmatrix}},\,{\begin{pmatrix}3\\7\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\0\\1\end{pmatrix}}$
is a basis as looked for. With respect to this basis, the linear mapping is described by the matrix
    ${\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}}.$

## Example

We want to apply fact to
    $M={\begin{pmatrix}0&3&7\\0&0&0\\0&0&0\end{pmatrix}}\,.$
We have
    $M^{2}=0\,.$
Therefore,
    $V_{1}=\operatorname {kern} M=\langle e_{1},7e_{2}-3e_{3}\rangle {\text{ and }}V_{2}=K^{3}.$
We have
    $V_{2}=V_{1}\oplus \langle e_{3}\rangle \,,$
so that we can choose
    $U_{2}=\langle e_{3}\rangle \,.$
We have
    $Me_{3}={\begin{pmatrix}7\\0\\0\end{pmatrix}}\in V_{1}\,.$
Therefore,
    $V_{1}=\langle {\begin{pmatrix}7\\0\\0\end{pmatrix}},7e_{2}-3e_{3}\rangle =U_{1}\,.$
Hence,
    ${\begin{pmatrix}0\\7\\-3\end{pmatrix}},\,{\begin{pmatrix}7\\0\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\0\\1\end{pmatrix}}$
is a looked-for basis. With respect to this basis, the linear mapping is described by the matrix
    ${\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}}.$
* ↑ Restriction as a mapping to $V$; the $U_{i}$ are in general not $\varphi$-invariant.