Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 27

In the last lecture, we have introduced generalized eigenspaces for an eigenvalue ${}\lambda$ of an endomorphism ${}\varphi$ as the kernel of ${}{\left(\varphi -\lambda \operatorname {Id} \right)}^{k}$ for a sufficient large exponent ${}k$. This means in particular that, if we restrict ${}\varphi -\lambda \operatorname {Id}$ to the corresponding generalized eigenspace, then a certain power of it is the zero mapping. Here, we study in general endomorphisms with the property that some power of it is zero.

Nilpotent mappings

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 27.3 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 Lemma 24.5 .

$\Box$

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 Lemma 27.7 and Theorem 25.10 .

$\Box$

The Jordan decomposition of a nilpotent endomorphism

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).   In the next lecture, we will apply these improvements to all generalized eigenspaces of a trigonalizable endomorphism, in order to achieve the so-called Jordan normal 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 Lemma 27.10 , 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 Lemma 27.11 .

$\Box$

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 Lemma 27.11 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 Lemma 27.11 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 Lemma 27.11 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

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

Footnotes

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