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

${\begin{pmatrix}0&c_{1}&0&\cdots &\cdots &0\\0&0&c_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&0&c_{n-2}&0\\0&\cdots &\cdots &0&0&c_{n-1}\\0&\cdots &\cdots &\cdots &0&0\end{pmatrix}},$

where ${}c_{i}$ equals ${}0$ or ${}1$. That is, ${}\varphi$ can be brought into Jordan normal form.

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

${\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}}.$