Modules over $k[x]$ and the Jordan Form

Introduction

A linear operator on a finite-dimensional vector space over a field $k$ is the same thing as a module over the polynomial ring $k[x]$: the indeterminate acts by the operator, and the module axioms are exactly the relations satisfied by iterates of the operator. Since $k[x]$ is a principal ideal domain, the structure theorem for finitely generated modules over a principal ideal domain applies and yields the rational canonical form, and when the polynomial invariant factors split into linear factors it yields the Jordan canonical form. The characteristic and minimal polynomials appear as the product of the invariant factors and as the largest one respectively.

Throughout, $k$ is a field, $V$ is a finite-dimensional $k$-vector space of dimension $n$, and $T:V \to V$ is $k$-linear. The structure theorem is that of the companion article of this category on modules over a principal ideal domain; the eigenvalues and diagonalisation of the article of this category on eigenvalues and diagonalisation are used as motivation and for the diagonalisability criterion, and the determinant is that of the article on the special linear group and the determinant.

The $k[x]$-Module of an Operator

Definition

Definition. For $p(x)=\sum_{i=0}^{d}a_ix^i \in k[x]$ and $v \in V$ put $p \cdot v=\sum_{i=0}^{d}a_iT^i(v)$, where $T^0=\operatorname{id}_V$. This makes $V$ a $k[x]$-module, and it is written $V_T$ when the operator needs to be named.

Proposition. (i) The construction gives $V$ the structure of a $k[x]$-module, and it is the unique $k[x]$-module structure with $x \cdot v=T(v)$ and $a \cdot v=av$ for $a \in k$.

(ii) The $k[x]$-submodules of $V_T$ are exactly the $T$-invariant $k$-subspaces of $V$.

(iii) The $k[x]$-module homomorphisms $V_T \to W_S$ are exactly the $k$-linear maps $f:V \to W$ with $fT=Sf$; in particular $V_T \cong V_S$ as $k[x]$-modules if and only if $T$ and $S$ are similar.

(iv) $V_T$ is a finitely generated torsion $k[x]$-module, and its annihilator is the ideal generated by the minimal polynomial $m_T$, the monic polynomial of least degree with $m_T(T)=0$.

Proof. (i) The only check is the compatibility $(pq)\cdot v=p\cdot(q\cdot v)$, which holds because $T^pT^q=T^{p+q}$ and scalars commute with $T$. (ii) A subset is a $k[x]$-submodule exactly when it is a $k$-subspace closed under $T$. (iii) A $k[x]$-linear map commutes with multiplication by $x$, that is, with $T$; conversely such an $f$ commutes with all polynomials in $T$. (iv) $V$ is generated as a $k[x]$-module by any $k$-basis, a finite set, and it is torsion because the minimal polynomial exists: $T$ satisfies its characteristic polynomial by Cayley–Hamilton, so the annihilator is nonzero.

Proposition. The $k[x]$-endomorphisms of $V_T$ are exactly the $k$-linear maps commuting with $T$. They always contain the subalgebra $k[T]=\{p(T):p \in k[x]\}$, with equality if and only if $T$ is cyclic, that is, if and only if the minimal polynomial of $T$ has degree $n$.

Proof. By (iii) with $W=V$ and $S=T$, a $k[x]$-endomorphism is a $k$-linear map commuting with $T$. Every $p(T)$ commutes with $T$, and if $T$ is cyclic with cyclic vector $v$ then a commuting $f$ is determined by $f(v)$, and $f(v)=p(T)v$ for some $p$ of degree $

The Structure Theorem Applied

Theorem. As a $k[x]$-module,

$$ V_T \;\cong\; k[x]/(f_1) \oplus k[x]/(f_2) \oplus \cdots \oplus k[x]/(f_k), $$

where $f_1 \mid f_2 \mid \cdots \mid f_k$ are monic polynomials of degree at least $1$. The $f_i$ are the invariant factors of $T$ and are uniquely determined by $T$. In this decomposition,

$$ \text{characteristic polynomial}=\prod_{i=1}^{k}f_i, \qquad \text{minimal polynomial}=f_k . $$

Proof. Since $k[x]$ is a principal ideal domain and $V_T$ is finitely generated and torsion, the structure theorem gives the displayed decomposition with the divisibility chain and uniqueness. The characteristic polynomial is multiplicative over direct sums, and the characteristic polynomial of $k[x]/(f)$ acting by multiplication by $x$ is $f$; multiplying gives $\prod_if_i$. The annihilator of the direct sum is the least common multiple of the annihilators $(f_i)$, which is $(f_k)$ because $f_i \mid f_k$ for all $i$.

Rational Canonical Form

The Companion Matrix

Definition. For a monic polynomial $f(x)=x^d+a_{d-1}x^{d-1}+\cdots+a_1x+a_0$, the companion matrix is the $d \times d$ matrix

$$ C(f)=\begin{pmatrix}0&0&\cdots&0&-a_0\\ 1&0&\cdots&0&-a_1\\ 0&1&\cdots&0&-a_2\\ \vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&\cdots&1&-a_{d-1}\end{pmatrix}. $$

Proposition. The $k[x]$-module $k[x]/(f)$ has $k$-basis $1,x,\dots,x^{d-1}$, on which multiplication by $x$ acts as $C(f)$; its characteristic and minimal polynomials are both $f$.

Proof. Multiplication by $x$ sends $x^j$ to $x^{j+1}$ for $j

Theorem (rational canonical form). Every operator $T$ on a finite-dimensional $k$-vector space is represented in some basis by the block diagonal matrix

$$ \operatorname{diag}\bigl(C(f_1),\dots,C(f_k)\bigr), $$

where $f_1 \mid \cdots \mid f_k$ are the invariant factors of $T$; this form is unique up to the ordering of the blocks, which is determined by the divisibility chain, and it requires no hypothesis on the field.

Proof. Combine the module decomposition with the proposition: in the basis $1,x,\dots,x^{\deg f_i-1}$ of each summand $k[x]/(f_i)$, the operator is $C(f_i)$. Uniqueness is the uniqueness of the invariant factors.

The rational canonical form is the form that exists over every field, including fields over which the invariant factors are irreducible of high degree; it is the form in which the module structure is visible with no splitting hypothesis.

Jordan Canonical Form

Splitting Invariant Factors

Definition. The characteristic polynomial splits over $k$ if it is a product of linear factors in $k[x]$, which is automatic when $k$ is algebraically closed. A Jordan block of size $e$ and eigenvalue $\lambda$ is the $e \times e$ matrix $J_e(\lambda)=\lambda I_e+N_e$, with $N_e$ the superdiagonal shift matrix having $1$s on the superdiagonal and $0$s elsewhere.

Proposition. If $f=(x-\lambda)^e$, then $k[x]/(f)$ has $k$-basis $(x-\lambda)^{e-1},\dots,(x-\lambda),1$ on which multiplication by $x$ acts as $J_e(\lambda)$, and its characteristic and minimal polynomials are both $(x-\lambda)^e$.

Proof. Multiplication by $x=\lambda+(x-\lambda)$ acts on $(x-\lambda)^j$ as $\lambda(x-\lambda)^j+(x-\lambda)^{j+1}$, so in the basis ordered from $(x-\lambda)^{e-1}$ down to $1$ the matrix is $\lambda I+N_e$; the polynomial statements are as for the companion matrix.

Theorem (Jordan canonical form). If the characteristic polynomial of $T$ splits over $k$, then $T$ is similar to a block diagonal matrix with Jordan blocks $J_{e}(\lambda)$, whose multiset of pairs $(e,\lambda)$ is uniquely determined by $T$. Equivalently, in the module decomposition each $k[x]/(f_i)$ is split by the Chinese remainder theorem into factors $k[x]/((x-\lambda)^{e})$.

Proof. The invariant factors $f_i$ factor into pairwise coprime prime powers $(x-\lambda)^e$; the Chinese remainder theorem gives $k[x]/(f_i) \cong \bigoplus_{\lambda,e}k[x]/((x-\lambda)^e)$, and each summand is a Jordan block by the proposition. Uniqueness is the uniqueness of the invariant factors together with the uniqueness of factorisation in $k[x]$.

Corollary. The number of Jordan blocks of eigenvalue $\lambda$ of size at least $j$ is $\operatorname{rk}((T-\lambda I)^{j-1})-\operatorname{rk}((T-\lambda I)^{j})$, and the number of blocks of size exactly $j$ is $\operatorname{rk}((T-\lambda I)^{j-1})-2\operatorname{rk}((T-\lambda I)^{j})+\operatorname{rk}((T-\lambda I)^{j+1})$.

Proof. On a Jordan block $J_e(\lambda)$, the rank of $(J_e(\lambda)-\lambda I)^m=N_e^m$ is $e-m$ for $m

These rank formulae are the computational content of the Jordan form: the size distribution of the blocks at an eigenvalue is read off from the ranks of the powers of $T-\lambda I$, with no need to compute eigenvectors explicitly.

The Minimal and Characteristic Polynomials

Definition. The characteristic polynomial of $T$ is $c_T(x)=\det(xI-T)$, and the minimal polynomial $m_T$ is the monic generator of the annihilator of $V_T$.

Theorem. (i) $c_T=\prod_{i=1}^kf_i$ and $m_T=f_k$, so $m_T \mid c_T$, and $c_T \mid m_T^{k}$; the two have the same roots in $k$.

(ii) (Cayley–Hamilton) $c_T(T)=0$.

(iii) $T$ is diagonalisable if and only if $m_T$ is a product of distinct linear factors over $k$, if and only if every invariant factor is squarefree and splits.

Proof. (i) is the theorem of the structure section, together with the observation that the roots of $c_T=\prod f_i$ are those of the $f_i$, all of which divide $f_k=m_T$, so the root sets agree; $c_T \mid m_T^k$ because each $f_i$ divides $f_k$. (ii) follows from $m_T \mid c_T$ together with (i). (iii) $T$ is diagonalisable exactly when $V_T$ is a direct sum of modules $k[x]/(x-\lambda)$, that is, when all invariant factors are products of distinct linear factors, which happens exactly when the largest one, $m_T$, has that form.

Worked Examples

Example. Over $k=\mathbb{R}$, let $T$ be the quarter-turn $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$. Its characteristic polynomial is $x^2+1$, irreducible over $\mathbb{R}$, so the only invariant factor is $x^2+1$, the minimal polynomial is $x^2+1$, and the rational canonical form is the companion matrix $C(x^2+1)=\begin{pmatrix}0&-1\\1&0\end{pmatrix}=T$ itself. There is no Jordan form over $\mathbb{R}$; over $\mathbb{C}$ the invariant factor splits as $(x-i)(x+i)$ and $T$ is diagonalisable with eigenvalues $\pm i$.

Example. Over any field, let $T=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$. Then $c_T=(x-\lambda)^2$ and $m_T=(x-\lambda)^2$, the invariant factor is $(x-\lambda)^2$, and $T=J_2(\lambda)$. The scalar $T=\lambda I$ has $c_T=(x-\lambda)^2$ and $m_T=x-\lambda$, invariant factors $(x-\lambda),(x-\lambda)$, and is already diagonal; these are the two similarity classes with characteristic polynomial $(x-\lambda)^2$, distinguished by the minimal polynomial. The rank formula gives $\operatorname{rk}(T-\lambda I)=1$ and $\operatorname{rk}((T-\lambda I)^2)=0$ for $J_2(\lambda)$, so there is one block of size $\ge 1$ and one of size $\ge 2$; for the scalar $\lambda I$ both ranks vanish, so there are two blocks of size $\ge 1$ and none of size $\ge 2$.

Example. Let $k$ be a field and let $T$ have invariant factors $f_1=(x-1)$ and $f_2=(x-1)(x-2)^2$. Then $c_T=(x-1)^2(x-2)^2$ and $m_T=(x-1)(x-2)^2$, and $T$ acts on a $4$-dimensional space, since $\deg f_1+\deg f_2=1+3=4$. The rational canonical form is $\operatorname{diag}\bigl(C(x-1),C((x-1)(x-2)^2)\bigr)$ with blocks of sizes $1$ and $3$. Splitting the second invariant factor by the Chinese remainder theorem, $k[x]/((x-1)(x-2)^2) \cong k[x]/(x-1) \oplus k[x]/((x-2)^2)$ because $x-1$ and $(x-2)^2$ are coprime, and the Jordan form is

$$ \operatorname{diag}\bigl(J_1(1),\,J_1(1),\,J_2(2)\bigr), $$

whose eigenvalue multiplicities are algebraic multiplicity $2$ at $\lambda=1$ and $2$ at $\lambda=2$, geometric multiplicity $2$ at $\lambda=1$ and $1$ at $\lambda=2$.

Example. Let $N$ be nilpotent on $k^n$ with invariant factors $x,x,\dots,x,x^3$, the last of degree $3$ and the remaining $n-3$ of degree $1$; then $c_N=x^n$ and $m_N=x^3$, and the Jordan form consists of one block $J_3(0)$ together with $n-3$ blocks $J_1(0)$. The rank formula returns exactly this: $\operatorname{rk}N=2$, $\operatorname{rk}N^2=1$ and $\operatorname{rk}N^3=0$, so there is one block of size at least $2$ and one of size at least $3$, and $n-2$ blocks of size at least $1$. The single condition $N^3=0 \neq N^2$ does not determine the type, since $J_3(0) \oplus J_3(0)$ also satisfies it and has rank $4$; it is the ranks, equivalently the invariant factors, that decide the block structure.

The Invariant Factors from $xI-T$

Theorem. Let $T$ be an $n \times n$ matrix over $k$ and regard $xI-T$ as a matrix over $k[x]$. Its Smith normal form has diagonal entries $s_1 \mid s_2 \mid \cdots \mid s_n$ with $s_i=1$ for $i \le n-k$ and $s_{n-k+j}=f_j$, where $f_1 \mid \cdots \mid f_k$ are the invariant factors of $T$. In particular $s_n=m_T$ and $\prod_is_i=\det(xI-T)=c_T$.

Proof. The matrix $xI-T$ is the matrix of multiplication by $x-T$ on $k[x]^n$, and its cokernel is $k[x]^n/(xI-T)k[x]^n \cong V_T$ as a $k[x]$-module. The structure theorem presents this cokernel with invariant factors the non-unit diagonal entries of the Smith normal form, and the entries are $1$ as long as the module needs fewer than $n$ generators; the divisibility chain is the normal form's. The product of the diagonal entries differs from $\det(xI-T)$ by a unit of $k[x]$, hence, both being monic, they are equal, and this product is $c_T$.

Example. For $T=J_2(\lambda)=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}$ one has $xI-T=\begin{pmatrix}x-\lambda&-1\\0&x-\lambda\end{pmatrix}$, whose entries have greatest common divisor $1$; the Smith normal form is therefore $\operatorname{diag}(1,(x-\lambda)^2)$, the invariant factor is $(x-\lambda)^2$, and the minimal and characteristic polynomials both equal $(x-\lambda)^2$. For $T=\lambda I_2$ the Smith normal form of $(x-\lambda)I_2$ is $\operatorname{diag}(x-\lambda,x-\lambda)$, giving invariant factors $(x-\lambda),(x-\lambda)$, minimal polynomial $x-\lambda$ and characteristic polynomial $(x-\lambda)^2$.

Summary

The structure of a linear operator on a finite-dimensional $k$-vector space is the structure of a finitely generated torsion module over the principal ideal domain $k[x]$, with $x$ acting as the operator. The module is a direct sum of cyclic modules $k[x]/(f_i)$ with $f_1 \mid \cdots \mid f_k$, the invariant factors; the characteristic polynomial is their product and the minimal polynomial is the largest one, so $m_T \mid c_T$, the two have the same roots, and Cayley–Hamilton holds. Submodules correspond to invariant subspaces, and $k[x]$-endomorphisms are the polynomials in $T$.

In the basis $1,x,\dots,x^{\deg f-1}$ of $k[x]/(f)$, multiplication by $x$ is the companion matrix $C(f)$, so the decomposition gives the rational canonical form $\operatorname{diag}(C(f_1),\dots,C(f_k))$, which exists over every field and is unique. When the invariant factors split into linear factors, each $k[x]/((x-\lambda)^e)$ contributes a Jordan block $J_e(\lambda)$ and one obtains the Jordan canonical form, unique up to ordering of blocks; the sizes of the blocks at an eigenvalue are computed from the ranks of the powers of $T-\lambda I$, by the difference formulae. The operator is diagonalisable exactly when the minimal polynomial is a product of distinct linear factors, equivalently when every invariant factor is squarefree and splits. The worked examples compute the rational and Jordan forms for a rotation, for the two classes with characteristic polynomial $(x-\lambda)^2$, for a mixed example with $(x-1)^2(x-2)^2$, and for a nilpotent operator.

Summary of Notation

Symbol	Meaning
$k$	a field
$V$	finite-dimensional $k$-vector space
$T \in \operatorname{End}_k(V)$	linear operator
$V_T$	$V$ as a $k[x]$-module
$m_T$, $c_T$	minimal and characteristic polynomials
$k[T]$	the centraliser algebra
$f_1 \mid \cdots \mid f_k$	invariant factors
$C(f)$	companion matrix of $f$
$J_e(\lambda)=\lambda I_e+N_e$	Jordan block
$N_e$	superdiagonal shift
$k[x]/(f)$	cyclic module with annihilator $(f)$
$\operatorname{rk}$	rank of an endomorphism