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

A decomposition theorem

Theorem

Let

$\phi :V⟶V$

denote a trigonalizable $K$-endomorphism on the finite-dimensional $K$-vector space $V$. Then there exists a decomposition

$\phi ={\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}+{\phi }_{\mathrm{n}\mathrm{i}\mathrm{l}},$

where ${\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}$ is diagonalizable and ${\phi }_{\mathrm{n}\mathrm{i}\mathrm{l}}$ is nilpotent, and where

${\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}\circ {\phi }_{\mathrm{n}\mathrm{i}\mathrm{l}}={\phi }_{\mathrm{n}\mathrm{i}\mathrm{l}}\circ {\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}$

holds.

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

$\phi ={\phi }_1\oplus \cdots \oplus {\phi }_m$

with ${\phi }_i=\phi {|}_{H_i}$. Let

$p_i:V⟶V$

denote the composition $V\to H_i\to V$, that is, $p_i$ is in particular a projection. We set

${\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}:={\lambda }_1{p}_1+\cdots +{\lambda }_m{p}_m.$

This mapping is obviously diagonalizable, on $H_i$ it is the multiplication with ${\lambda }_i$. Sei

${\phi }_{\mathrm{n}\mathrm{i}\mathrm{l}}:=\phi -{\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}.$

The property of this mapping of being nilpotent can be checked on the $H_i$ separately. There, we have

$(\phi -{\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}){|}_{H_i}={\phi }_i-\left({\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}\right){|}_{H_i}={\phi }_i-{\lambda }_i{\mathrm{Id}}_{H_i},$

so this is nilpotent. Moreover, ${\phi }_j$ and $p_i$ commute, since $p_i$ induces the identity on $H_i$ and on $H_j$, $j\ne i$, the zero mapping. Therefore, also the direct sums of those commute, and hence also $\phi$ and ${\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}$ commute. Thus, ${\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}$ and $\phi -{\phi }_{\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{g}}={\phi }_{\mathrm{n}\mathrm{i}\mathrm{l}}$ commute.

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Under the conditions given in the theorem, this decomposition is even unique.

Definition

An endomorphism

$\phi :V⟶V$

on a $K$-vector space is called unipotent, if we can write

$\phi ={\mathrm{Id}}_V+\psi$

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

$\left(\begin{array}{cccccc}\lambda & 1& 0& \cdots & \cdots & 0\\ 0& \lambda & 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& \lambda & 1& 0\\ 0& \cdots & \cdots & 0& \lambda & 1\\ 0& \cdots & \cdots & \cdots & 0& \lambda \end{array}\right).$

If we consider such a Jordan matrix as a linear mapping $\phi$ on the standard space $K^n$, then

$\phi (e_1)=\lambda e_1\text{ and }\phi (e_k)=\lambda e_k+e_{k-1}\text{ for all }k\ge 2.$

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

$e_{k-1}=(\phi -\lambda \cdot \mathrm{Id})(e_k)$

for $k\ge 2$. The eigenvector $e_1$ is a generating element of the kernel of the mapping $\psi :=\phi -\lambda \mathrm{Id}$, and the other standard vectors $e_k$ arise successively as preimages of $e_{k-1}$ under $\psi$.

Definition

A square matrix of the form

$\left(\begin{array}{ccccc}{J}_1& 0& \cdots & \cdots & 0\\ 0& J_2& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& J_{k-1}& 0\\ 0& \cdots & \cdots & 0& J_k\end{array}\right),$

where each $J_i$ is a Jordan matrix, is called a matrix in

Jordan normal form.

The Jordan matrices occurring here are called the Jordan blocks of the matrix. Their eigenvalues might be different or equal. In the matrix

$\left(\begin{array}{cccccc}2& 1& 0& 0& 0& 0\\ 0& 2& 0& 0& 0& 0\\ 0& 0& 4& 1& 0& 0\\ 0& 0& 0& 4& 1& 0\\ 0& 0& 0& 0& 4& 0\\ 0& 0& 0& 0& 0& 2\end{array}\right),$

there are three Jordan blocks,

$\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right),\left(\begin{array}{ccc}4& 1& 0\\ 0& 4& 1\\ 0& 0& 4\end{array}\right)\text{ and }\left(\begin{array}{c}2\end{array}\right)$

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

$\phi :V⟶V$

on a finite-dimensional $K$-vector space $V$, there exists a basis, such that the describing matrix is in

Jordan normal form.

Proof

Since $\phi$ is trigonalizable, we can apply Lemma 26.14 . Hence, there exists a direct sum decomposition

$V={\mathrm{GeEig}}_{{\lambda }_1}(\phi )\oplus \cdots \oplus {\mathrm{GeEig}}_{{\lambda }_m}(\phi ),$

where the generalized eigenspaces are $\phi$-invariant. Looking at the situation for each generalized eigenspace, we may assume that $\phi$ has only one eigenvalue $\lambda$, and that

$V={\mathrm{GeEig}}_{\lambda }(\phi )$

holds. Then,

$\psi =\phi -\lambda {\mathrm{Id}}_V$

is nilpotent. Therefore, because of Fact *****, there exists a basis such that $\psi$ is described by a matrix of the form

$\left(\begin{array}{cccccc}0& c_1& 0& \cdots & \cdots & 0\\ 0& 0& c_2& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& 0& c_{n-2}& 0\\ 0& \cdots & \cdots & 0& 0& c_{n-1}\\ 0& \cdots & \cdots & \cdots & 0& 0\end{array}\right),$

where the $c_i$ equal $0$ or equal $1$. With respect to this basis,

$\phi =\psi +\lambda {\mathrm{Id}}_V$

has the form

$\left(\begin{array}{cccccc}\lambda & c_1& 0& \cdots & \cdots & 0\\ 0& \lambda & c_2& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& \lambda & c_{n-2}& 0\\ 0& \cdots & \cdots & 0& \lambda & c_{n-1}\\ 0& \cdots & \cdots & \cdots & 0& \lambda \end{array}\right).$

$◻$

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 $\phi :V\to V$ 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

$\mathrm{kern}{(\phi -\lambda \mathrm{Id})}^s=\mathrm{kern}{(\phi -\lambda \mathrm{Id})}^{s+1}.$

This kernel is the generalized eigenspace to $\lambda$. We set

$V_i=\mathrm{kern}{(\phi -\lambda \mathrm{Id})}^i\subseteq {\mathrm{GeEig}}_{\lambda }(\phi )$

for $i=1,\dots ,s$. This yields the chain

$V_1={\mathrm{Eig}}_{}\left(\lambda \right)\subseteq V_2\subset \cdots \subset V_{s-1}\subset V_s={\mathrm{GeEig}}_{\lambda }(\phi ).$

Now, we choose a vector $u$ from $V_s\setminus V_{s-1}$. The vectors

$u,(\phi -\lambda \mathrm{Id})(u),(\phi -\lambda \mathrm{Id}{)}^2(u),\dots ,(\phi -\lambda \mathrm{Id}{)}^{s-1}(u)$

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 $\phi -\lambda {\mathrm{Id}}_V$ of the vectors, that is, we have to solve the equation

$v=(\phi -\lambda {\mathrm{Id}}_V)(v^{\prime }),$

then

$v^{\prime }=(\phi -\lambda {\mathrm{Id}}_V)(v^{\prime \prime }),$

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 $\phi -\lambda {\mathrm{Id}}_V$.

Example

We consider the matrix

$M=\left(\begin{array}{ccc}2& 2& 1\\ 0& 2& 3\\ 0& 0& 2\end{array}\right),$

and want to bring it to Jordan normal form. bringen. The vector $u=e_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ is an eigenvector to the eigenvalue $2$. We have

$A:=M-2E_3=\left(\begin{array}{ccc}0& 2& 1\\ 0& 0& 3\\ 0& 0& 0\end{array}\right),$

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 $v=\left(\begin{array}{c}0\\ \frac{1}{2}\\ 0\end{array}\right)$. Finally, we need a solution for $v=Aw$. This yields the equation $w=\left(\begin{array}{c}0\\ -\frac{1}{12}\\ \frac{1}{6}\end{array}\right)$. 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

$\left(\begin{array}{ccc}2& 1& 0\\ 0& 2& 1\\ 0& 0& 2\end{array}\right).$

This matrix is a Jordan matrix and, in particular, in Jordan normal form.

Example

We consider the matrix

$M=\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 3\\ 0& 0& 2\end{array}\right)$

and want to bring it to Jordan normal form. The vectors $u=e_1=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ and $v=e_2=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$ are linearly independent eigenvectors to the eigenvalue $2$. We have

$A:=M-2E_3=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 3\\ 0& 0& 0\end{array}\right),$

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 $w=\left(\begin{array}{c}0\\ 0\\ \frac{1}{3}\end{array}\right)$. 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

$\left(\begin{array}{ccc}2& 0& 0\\ 0& 2& 1\\ 0& 0& 2\end{array}\right).$

This matrix is in Jordan normal form with the Jordan blocks $\left(\begin{array}{c}2\end{array}\right)$ and $\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right)$.

Example

We consider the matrix

$M=\left(\begin{array}{cccc}3& 1& 0& 4\\ 0& -1& 2& 1\\ 0& 0& -1& 0\\ 0& 0& 0& 3\end{array}\right)$

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

$M-3E_4=\left(\begin{array}{cccc}0& 1& 0& 4\\ 0& -4& 2& 1\\ 0& 0& -4& 0\\ 0& 0& 0& 0\end{array}\right),$

therefore, $\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right)$ 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

$\begin{array}{cc}{\left(\begin{array}{cccc}0& 1& 0& 4\\ 0& -4& 2& 1\\ 0& 0& -4& 0\\ 0& 0& 0& 0\end{array}\right)}^2& =\left(\begin{array}{cccc}0& 1& 0& 4\\ 0& -4& 2& 1\\ 0& 0& -4& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{cccc}0& 1& 0& 4\\ 0& -4& 2& 1\\ 0& 0& -4& 0\\ 0& 0& 0& 0\end{array}\right)\\ & =\left(\begin{array}{cccc}0& -4& 2& 1\\ 0& 16& -16& -4\\ 0& 0& 16& 0\\ 0& 0& 0& 0\end{array}\right),\end{array}$

a new element of the kernel is $\left(\begin{array}{c}0\\ 1\\ 0\\ 4\end{array}\right)$. Thus, we have

${\mathrm{GeEig}}_3(M)=⟨\left(\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\\ 4\end{array}\right)⟩.$

Because of

$\left(\begin{array}{cccc}0& 1& 0& 4\\ 0& -4& 2& 1\\ 0& 0& -4& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}0\\ 1\\ 0\\ 4\end{array}\right)=\left(\begin{array}{c}17\\ 0\\ 0\\ 0\end{array}\right),$

we can use the vectors $\left(\begin{array}{c}17\\ 0\\ 0\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\\ 0\\ 4\end{array}\right)$ to establish the first Jordan block.

We have

$M+1E_4=\left(\begin{array}{cccc}4& 1& 0& 4\\ 0& 0& 2& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 4\end{array}\right),$

therefore, $\left(\begin{array}{c}1\\ -4\\ 0\\ 0\end{array}\right)$ belongs to the kernel. The rank of this matrix is again $3$, and the kernel has dimension one. The second power is

$\begin{array}{cc}{\left(\begin{array}{cccc}4& 1& 0& 4\\ 0& 0& 2& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 4\end{array}\right)}^2& =\left(\begin{array}{cccc}4& 1& 0& 4\\ 0& 0& 2& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 4\end{array}\right)\left(\begin{array}{cccc}4& 1& 0& 4\\ 0& 0& 2& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 4\end{array}\right)\\ & =\left(\begin{array}{cccc}16& 4& 2& 33\\ 0& 0& 0& 4\\ 0& 0& 0& 0\\ 0& 0& 0& 16\end{array}\right),\end{array}$

a new element in the kernel is $\left(\begin{array}{c}0\\ 1\\ -2\\ 0\end{array}\right)$. Thus, we have

${\mathrm{GeEig}}_{-1}(M)=⟨\left(\begin{array}{c}1\\ -4\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ -2\\ 0\end{array}\right)⟩.$

Because of

$\left(\begin{array}{cccc}4& 1& 0& 4\\ 0& 0& 2& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 4\end{array}\right)\left(\begin{array}{c}0\\ 1\\ -2\\ 0\end{array}\right)=\left(\begin{array}{c}1\\ -4\\ 0\\ 0\end{array}\right),$

we can use the vectors $\left(\begin{array}{c}1\\ -4\\ 0\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\\ -2\\ 0\end{array}\right)$ to establish the second Jordan block. Altogether, the linear mapping defined by $M$ has, with respect the basis

$\left(\begin{array}{c}17\\ 0\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\\ 4\end{array}\right),\left(\begin{array}{c}1\\ -4\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ -2\\ 0\end{array}\right),$

the Jordan normal form

$\left(\begin{array}{cccc}3& 1& 0& 0\\ 0& 3& 0& 0\\ 0& 0& -1& 1\\ 0& 0& 0& -1\end{array}\right).$

Endomorphisms with finite order

In Lemma 24.11 , we have seen that permutation matrices over $ℂ$ are diagonalizable. This is true over $ℂ$ for all endomorphisms of finite order.

Lemma

Every invertible matrix $M\in {\mathrm{GL}}_n\left(ℂ\right)$ of finite order is

diagonalizable.

Proof

The matrix is trigonalizable and can be brought, due to Theorem 28.5 , into Jordan normal form. We show that the Jordan blocks

$\left(\begin{array}{cccccc}\lambda & 1& 0& \cdots & \cdots & 0\\ 0& \lambda & 1& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& \lambda & 1& 0\\ 0& \cdots & \cdots & 0& \lambda & 1\\ 0& \cdots & \cdots & \cdots & 0& \lambda \end{array}\right)$

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

$\left(\begin{array}{cccccc}1& a& 0& \cdots & \cdots & 0\\ 0& 1& a& 0& \cdots & 0\\ ⋮& \ddots & \ddots & \ddots & \ddots & ⋮\\ 0& \cdots & 0& 1& a& 0\\ 0& \cdots & \cdots & 0& 1& a\\ 0& \cdots & \cdots & \cdots & 0& 1\end{array}\right)$

(with $a\ne 0$). 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.

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Footnotes

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