# Linear algebra (Osnabrück 2024-2025)/Part I/Lecture 28

    A decomposition theorem

## Theorem

Let
    $\varphi \colon V\longrightarrow V$
denote a trigonalizable $K$-endomorphism on the finite-dimensional $K$-vector space $V$. Then there exists a decomposition
    $\varphi =\varphi _{\rm {diag}}+\varphi _{\rm {nil}}\,,$
where $\varphi _{\rm {diag}}$ is diagonalizable and $\varphi _{\rm {nil}}$ is nilpotent, and where
    $\varphi _{\rm {diag}}\circ \varphi _{\rm {nil}}=\varphi _{\rm {nil}}\circ \varphi _{\rm {diag}}\,$

### Proof

Due to Lemma 26.14 , we have
    $V=H_{1}\oplus \cdots \oplus H_{m}\,,$
where the $H_{i}$ are the generalized eigenspaces for the eigenvalues $\lambda _{i}$, and we have
    $\varphi =\varphi _{1}\oplus \cdots \oplus \varphi _{m}\,$
with $\varphi _{i}=\varphi {|}_{H_{i}}$. Let
    $p_{i}\colon V\longrightarrow V$
denote the composition $V\rightarrow H_{i}\rightarrow V$, that is, $p_{i}$ is in particular a projection. We set
    $\varphi _{\rm {diag}}:=\lambda _{1}p_{1}+\cdots +\lambda _{m}p_{m}\,.$
This mapping is obviously diagonalizable, on $H_{i}$ it is the multiplication with $\lambda _{i}$. Sei
    $\varphi _{\rm {nil}}:=\varphi -\varphi _{\rm {diag}}\,.$
The property of this mapping of being nilpotent can be checked on the $H_{i}$ separately. There, we have
    ${\left(\varphi -\varphi _{\rm {diag}}\right)}{|}_{H_{i}}=\varphi _{i}-{\left(\varphi _{\rm {diag}}\right)}{|}_{H_{i}}=\varphi _{i}-\lambda _{i}\operatorname {Id} _{H_{i}}\,,$
so this is nilpotent. Moreover, $\varphi _{j}$ and $p_{i}$ commute, since $p_{i}$ induces the identity on $H_{i}$ and on $H_{j}$, $j\neq i$, the zero mapping. Therefore, also the direct sums of those commute, and hence also $\varphi$ and $\varphi _{\rm {diag}}$ commute. Thus, $\varphi _{\rm {diag}}$ and $\varphi -\varphi _{\rm {diag}}=\varphi _{\rm {nil}}$ commute.
$\Box$
Under the conditions given in the theorem, this decomposition is even unique.

## Definition

An endomorphism
    $\varphi \colon V\longrightarrow V$
on a $K$-vector space is called unipotent, if we can write
    $\varphi =\operatorname {Id} _{V}+\psi \,$
with some nilpotent mapping
$\psi$
For a unipotent mapping, the diagonalizable part in the sense of the canonical decomposition from Theorem 28.1
is simple, it is just the identity.
    Jordan normal form

## Definition

Let $K$ be a field and $\lambda \in K$. A Jordan matrix (to the eigenvalue $\lambda$) is a square matrix of the form
    ${\begin{pmatrix}\lambda &1&0&\cdots &\cdots &0\\0&\lambda &1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda &1&0\\0&\cdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &\cdots &0&\lambda \end{pmatrix}}.$
If we consider such a Jordan matrix as a linear mapping $\varphi$ on the standard space $K^{n}$, then
    $\varphi (e_{1})=\lambda e_{1}{\text{ and }}\varphi (e_{k})=\lambda e_{k}+e_{k-1}{\text{ for all }}k\geq 2.$
In particular, $e_{1}$ is a eigenvector to the eigenvalue $\lambda$. A simple observation shows that there is no further eigenvector linearly independent to $e_{1}$ (see Exercise 28.22 ). The property on the right is equivalent with the condition
    $e_{k-1}=(\varphi -\lambda \cdot \operatorname {Id} )(e_{k})\,$
for $k\geq 2$. The eigenvector $e_{1}$ is a generating element of the kernel of the mapping $\psi :=\varphi -\lambda \operatorname {Id}$, and the other standard vectors $e_{k}$ arise successively as preimages of $e_{k-1}$ under $\psi$.

## Definition

A square matrix of the form
    ${\begin{pmatrix}J_{1}&0&\cdots &\cdots &0\\0&J_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&J_{k-1}&0\\0&\cdots &\cdots &0&J_{k}\end{pmatrix}},$
where each $J_{i}$ is a Jordan matrix, is called a matrix in
The Jordan matrices occurring here are called the Jordan blocks of the matrix. Their eigenvalues might be different or equal. In the matrix
    ${\begin{pmatrix}2&1&0&0&0&0\\0&2&0&0&0&0\\0&0&4&1&0&0\\0&0&0&4&1&0\\0&0&0&0&4&0\\0&0&0&0&0&2\end{pmatrix}},$
there are three Jordan blocks,
    ${\begin{pmatrix}2&1\\0&2\end{pmatrix}},\,{\begin{pmatrix}4&1&0\\0&4&1\\0&0&4\end{pmatrix}}{\text{ and }}{\begin{pmatrix}2\end{pmatrix}}$
with eigenvalues $2,4$ and again $2$.
We state and prove now the theorem about the Jordan normal form for trigonalizable endomorphisms.

## Theorem

For every trigonalizable endomorphism
    $\varphi \colon V\longrightarrow V$
on a finite-dimensional $K$-vector space $V$, there exists a basis, such that the describing matrix is in

### Proof

Since $\varphi$ is trigonalizable, we can apply Lemma 26.14 . Hence, there exists a direct sum decomposition
    $V=\operatorname {GeEig} _{\lambda _{1}}(\varphi )\oplus \cdots \oplus \operatorname {GeEig} _{\lambda _{m}}(\varphi )\,,$
where the generalized eigenspaces are $\varphi$-invariant. Looking at the situation for each generalized eigenspace, we may assume that $\varphi$ has only one eigenvalue $\lambda$, and that
    $V=\operatorname {GeEig} _{\lambda }(\varphi )\,$
holds. Then,
    $\psi =\varphi -\lambda \operatorname {Id} _{V}\,$
is nilpotent. Therefore, because of Fact *****, there exists a basis such that $\psi$ is described by a matrix of 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 the $c_{i}$ equal $0$ or equal $1$. With respect to this basis,
    $\varphi =\psi +\lambda \operatorname {Id} _{V}\,$
has the form
    ${\begin{pmatrix}\lambda &c_{1}&0&\cdots &\cdots &0\\0&\lambda &c_{2}&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda &c_{n-2}&0\\0&\cdots &\cdots &0&\lambda &c_{n-1}\\0&\cdots &\cdots &\cdots &0&\lambda \end{pmatrix}}.$
$\Box$
Hence, every upper triangular matrix is similar to a matrix in Jordan normal form. Over the complex numbers, every matrix can be brought to Jordan normal form. If a matrix has Jordan normal form, then we can read of directly the diagonalizable and the nilpotent part in the sense of Theorem 28.1 : The diagonal yields the diagonalizable part, and the entries which are strictly above the diagonal, yield the nilpotent part (in general, this is not true for upper triangular matrices).

## Method

We describe how to find to a linear trigonalizable mapping $\varphi \colon V\rightarrow V$ a basis, such that describing matrix with respect to this basis is in Jordan normal form. For this, we determine, for every eigenvalue $\lambda \in K$, the minimal exponent $s$ with
    $\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{s}=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{s+1}\,.$
This kernel is the generalized eigenspace to $\lambda$. We set
    $V_{i}=\operatorname {kern} {\left(\varphi -\lambda \operatorname {Id} \right)}^{i}\subseteq \operatorname {GeEig} _{\lambda }(\varphi )\,$
for $i=1,\ldots ,s$. This yields the chain
    $V_{1}=\operatorname {Eig} _{\left(\lambda \right)}\subseteq V_{2}\subset \cdots \subset V_{s-1}\subset V_{s}=\operatorname {GeEig} _{\lambda }(\varphi )\,.$
Now, we choose a vector $u$ from $V_{s}\setminus V_{s-1}$. The vectors
    $u,(\varphi -\lambda \operatorname {Id} )(u),(\varphi -\lambda \operatorname {Id} )^{2}(u),\ldots ,(\varphi -\lambda \operatorname {Id} )^{s-1}(u)$
form a basis for a Jordan block. If this basis generates the generalized eigenspace, then we are done. Otherwise, we look in $V_{s}\setminus V_{s-1}$ for another vector which is linearly independent to $u$ and to $V_{s-1}$. Again, we add this vector and all its successive images. If $V_{s}\setminus V_{s-1}$ is exhausted, then we look whether $V_{s-1}\setminus V_{s-2}$ is already covered, and so on. If the generalized eigenspace to $\lambda$ is covered, then we continue with the next eigenvalue.
Under certain circumstances, we can also start with a basis of the eigenspace. If, for example, the eigenspace to $\lambda$ is one-dimensional, then we can choose an eigenvector $v$ for $\lambda$, and we can find successively preimages under $\varphi -\lambda \operatorname {Id} _{V}$ of the vectors, that is, we have to solve the equation
    $v={\left(\varphi -\lambda \operatorname {Id} _{V}\right)}(v')\,,$
then
    $v'={\left(\varphi -\lambda \operatorname {Id} _{V}\right)}(v^{\prime \prime })\,,$
etc.
If, for example, the eigenspace is $k$-dimensional and the generalized eigenspace is $(k+1)$-dimensional, then we only have to find a preimage for one eigenvector under $\varphi -\lambda \operatorname {Id} _{V}$.

## Example

We consider the matrix
    $M={\begin{pmatrix}2&2&1\\0&2&3\\0&0&2\end{pmatrix}}\,,$
and want to bring it to Jordan normal form. bringen. The vector $u=e_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$ is an eigenvector to the eigenvalue $2$. We have
    $A:=M-2E_{3}={\begin{pmatrix}0&2&1\\0&0&3\\0&0&0\end{pmatrix}}\,,$
so that there exists no further linearly independent eigenvector. We look at the linear system $e_{1}=Av$. This imples (looking at the second row) $v_{3}=0$ and so $2v_{2}=1$ (we can choose $v_{1}$ freely to be $0$). Hence, we set $v={\begin{pmatrix}0\\{\frac {1}{2}}\\0\end{pmatrix}}$. Finally, we need a solution for $v=Aw$. This yields the equation $w={\begin{pmatrix}0\\-{\frac {1}{12}}\\{\frac {1}{6}}\end{pmatrix}}$. The matrix $M$ acts as
    $Mu=2u,\,Mv=2v+u,\,Mw=2w+v,$
so that the mapping is described with respect to the basis $u,v,w$ by
    ${\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}}.$
This matrix is a Jordan matrix and, in particular, in Jordan normal form.

## Example

We consider the matrix
    $M={\begin{pmatrix}2&0&0\\0&2&3\\0&0&2\end{pmatrix}}\,$
and want to bring it to Jordan normal form. The vectors $u=e_{1}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$ and $v=e_{2}={\begin{pmatrix}0\\1\\0\end{pmatrix}}$ are linearly independent eigenvectors to the eigenvalue $2$. We have
    $A:=M-2E_{3}={\begin{pmatrix}0&0&0\\0&0&3\\0&0&0\end{pmatrix}}\,,$
so that $u$ and $v$ span this eigenspace. An eigenvector must be the image of some vector under the matrix $A$. In fact, the linear system
    $e_{2}=Aw\,$
has the solution $w={\begin{pmatrix}0\\0\\{\frac {1}{3}}\end{pmatrix}}$. Therefore, the matrix $M$ acts in the following way
    $Mu=2u,\,Mv=2v,\,Mw=2w+v.$
Hence, the mapping is described, with respect to the basis $u,v,w$, by the matrix
    ${\begin{pmatrix}2&0&0\\0&2&1\\0&0&2\end{pmatrix}}.$
This matrix is in Jordan normal form with the Jordan blocks ${\begin{pmatrix}2\end{pmatrix}}$ and ${\begin{pmatrix}2&1\\0&2\end{pmatrix}}$.

## Example

We consider the matrix
    $M={\begin{pmatrix}3&1&0&4\\0&-1&2&1\\0&0&-1&0\\0&0&0&3\end{pmatrix}}\,$
and want to bring in in Jordan normal for. Here, we have two eigenvalues and, therefore, two two-dimensional generalized eigenspaces, which we treat separately. We have
    $M-3E_{4}={\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}\,,$
therefore, ${\begin{pmatrix}1\\0\\0\\0\end{pmatrix}}$ belongs to the kernel. The determinant of the upper right submatrix is not $0$, so the rank of the matrix is $3$, and its kernel is one-dimensional. The second power is
    ${\begin{aligned}{\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}^{2}&={\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}{\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}\\&={\begin{pmatrix}0&-4&2&1\\0&16&-16&-4\\0&0&16&0\\0&0&0&0\end{pmatrix}},\end{aligned}}$
a new element of the kernel is ${\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}$. Thus, we have
    $\operatorname {GeEig} _{3}(M)=\langle {\begin{pmatrix}1\\0\\0\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}\rangle \,.$
Because of
    ${\begin{pmatrix}0&1&0&4\\0&-4&2&1\\0&0&-4&0\\0&0&0&0\end{pmatrix}}{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}={\begin{pmatrix}17\\0\\0\\0\end{pmatrix}}\,,$
we can use the vectors ${\begin{pmatrix}17\\0\\0\\0\end{pmatrix}}$ and ${\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}$ to establish the first Jordan block.
We have
    $M+1E_{4}={\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}\,,$
therefore, ${\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}$ belongs to the kernel. The rank of this matrix is again $3$, and the kernel has dimension one. The second power is
    ${\begin{aligned}{\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}^{2}&={\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}{\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}\\&={\begin{pmatrix}16&4&2&33\\0&0&0&4\\0&0&0&0\\0&0&0&16\end{pmatrix}},\end{aligned}}$
a new element in the kernel is ${\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}$. Thus, we have
    $\operatorname {GeEig} _{-1}(M)=\langle {\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}\rangle \,.$
Because of
    ${\begin{pmatrix}4&1&0&4\\0&0&2&1\\0&0&0&0\\0&0&0&4\end{pmatrix}}{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}={\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}\,,$
we can use the vectors ${\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}$ and ${\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}}$ to establish the second Jordan block. Altogether, the linear mapping defined by $M$ has, with respect the basis
    ${\begin{pmatrix}17\\0\\0\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\1\\0\\4\end{pmatrix}}\,,{\begin{pmatrix}1\\-4\\0\\0\end{pmatrix}}\,,{\begin{pmatrix}0\\1\\-2\\0\end{pmatrix}},$
the Jordan normal form
    ${\begin{pmatrix}3&1&0&0\\0&3&0&0\\0&0&-1&1\\0&0&0&-1\end{pmatrix}}.$
    Endomorphisms with finite order
In Lemma 24.11 , we have seen that permutation matrices over $\mathbb {C}$ are diagonalizable. This is true over $\mathbb {C}$ for all endomorphisms of finite order.

## Lemma

Every invertible matrix $M\in \operatorname {GL} _{n}\!{\left(\mathbb {C} \right)}$ of finite order is

### Proof

The matrix is trigonalizable and can be brought, due to Theorem 28.5 , into Jordan normal form. We show that the
Jordan blocks
    ${\begin{pmatrix}\lambda &1&0&\cdots &\cdots &0\\0&\lambda &1&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&\lambda &1&0\\0&\cdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &\cdots &0&\lambda \end{pmatrix}}$
are trivial. Because of the finite order, $\lambda$ is a root of unity. By multiplying with $\lambda ^{-1}E_{n}$, we can assume that we have a matrix of the form
    ${\begin{pmatrix}1&a&0&\cdots &\cdots &0\\0&1&a&0&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\0&\cdots &0&1&a&0\\0&\cdots &\cdots &0&1&a\\0&\cdots &\cdots &\cdots &0&1\end{pmatrix}}$
(with $a\neq 0$). If this is not an $1\times 1$-matrix, then there exist two vectors $u,v$, where $u$ is an eigenvector and where $v$ is sent to $v+au$. The $k$-th iteration of the matrix sends $v$ to $v+kau$, and this is never $v$, contradicting the property of being of finite order.
$\Box$
    Footnotes
* ↑ Some authors define a Jordan matrix one where the $1$ are below the diagonal.
* ↑ In the context of trigonalizable mappings and in order to find the Jordan normal form, it is useful to work with $\varphi -\lambda \cdot \operatorname {Id}$ instead of $\lambda \cdot \operatorname {Id} -\varphi$.