---
title: Jordan Canonical Form
module: Modules over PIDs and Canonical Forms
moduleNumber: 9
lessonNumber: 3
order: 903
summary: >
  When the base field contains all the eigenvalues, the elementary divisors of an
  operator are powers of linear polynomials, and each cyclic summand becomes a
  Jordan block: an eigenvalue on the diagonal with ones just above it. Stacking the
  blocks gives the Jordan canonical form, unique up to reordering, as close to
  diagonal as the operator allows. Diagonalizability reads off the minimal
  polynomial, and the block sizes are counted by ranks of powers of the operator
  minus the eigenvalue.
topics: [Modules over PIDs and Canonical Forms]
draft: false
sources:
  - book: Dummit & Foote
    ref: "Ch. 12 — Modules over Principal Ideal Domains; §12.3 The Jordan Canonical Form"
---

The [rational canonical form](/abstract-algebra/modules-over-pids/rational-canonical-form)
uses the invariant-factor decomposition of the $F[x]$-module $V_T$. The
[elementary-divisor decomposition](/abstract-algebra/modules-over-pids/structure-theorem-over-pids)
of the same module gives a different matrix — one that is nearly diagonal, at the
cost of needing the eigenvalues to lie in the field.

Throughout, $V$ is a finite-dimensional vector space over $F$, $T : V \to V$ is a
linear operator, $V_T$ is $V$ made into an $F[x]$-module by $x \cdot v = T(v)$, and
$A$ is a matrix of $T$.

## Splitting into linear prime powers

The elementary divisors of $V_T$ are the prime-power factors of its invariant
factors. Over $F[x]$ the primes are the monic irreducible polynomials, so an
elementary divisor is $p(x)^k$ for some irreducible $p$. To make it as simple as
possible, assume every irreducible factor is linear.

> **Assumption ($F$ contains the eigenvalues).** Every invariant factor of $T$
> factors into linear polynomials over $F$. Equivalently, since the product of the
> elementary divisors is the characteristic polynomial, $F$ contains all
> eigenvalues of $T$. This holds automatically when $F$ is algebraically closed,
> for instance $F = \mathbb{C}$.

Under this assumption the elementary divisors are powers $(x - \lambda)^k$ with
$\lambda \in F$ an eigenvalue, and the structure theorem gives
$$
V_T \;\cong\; \bigoplus_i F[x] \big/ \bigl((x - \lambda_i)^{k_i}\bigr).
$$
Each summand is a cyclic $F[x]$-module, and the task is to choose a good basis for
it.

## The Jordan block

Fix one summand $F[x]/((x - \lambda)^k)$. Instead of the coefficient basis $1, x,
\dots, x^{k-1}$ used for companion matrices, take the basis built from powers of
$(x - \lambda)$:
$$
(x - \lambda)^{k-1},\; (x - \lambda)^{k-2},\; \dots,\; (x - \lambda),\; 1.
$$
This is a basis because the change from the coefficient basis is upper triangular
with $1$'s on the diagonal. Writing $x = \lambda + (x - \lambda)$ and using
$(x - \lambda)^k = 0$ in the quotient, multiplication by $x$ acts as
$$
(x - \lambda)^{j} \;\longmapsto\; \lambda\,(x - \lambda)^{j} + (x - \lambda)^{j+1},
$$
so on each basis vector $T$ returns $\lambda$ times itself plus the next vector up.
The matrix is diagonal with an extra superdiagonal of ones.

> **Definition (Jordan block).** The $k \times k$ **Jordan block** with eigenvalue
> $\lambda$ is
> $$
> J_k(\lambda) =
> \begin{pmatrix}
> \lambda & 1 & & \\
> & \lambda & 1 & \\
> & & \ddots & \ddots \\
> & & & \lambda & 1 \\
> & & & & \lambda
> \end{pmatrix},
> $$
> with $\lambda$ on the main diagonal, $1$'s on the first superdiagonal, and zeros
> elsewhere. Its characteristic and minimal polynomials are both $(x-\lambda)^k$.

$$
% caption: A single Jordan block of size $k$: the eigenvalue (here drawn as $d$ for $\lambda$) repeated on the diagonal, ones on the first superdiagonal, zeros elsewhere.
\begin{tikzpicture}[font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) rectangle (4,4);
\foreach \i in {0,1,2,3} \node[fill=acc!12, draw=acc, minimum size=8mm, inner sep=0pt] at ({\i+0.5},{3.5-\i}) {$d$};
\foreach \i in {0,1,2} \node[minimum size=8mm, inner sep=0pt, acc] at ({\i+1.5},{3.5-\i}) {$1$};
\node[black] at (0.5,0.5) {$0$};
\node[black] at (3.5,3.5) {$0$};
\node[black, anchor=south, font=\footnotesize] at (2,4.1) {diagonal: eigenvalue $d$};
\node[acc, anchor=west, font=\footnotesize] at (4.3,2.3) {superdiagonal ones};
\draw[acc, ->] (4.25,2.3) -- (2.7,2.9);
\end{tikzpicture}
$$

A Jordan block is the sum of the scalar $\lambda I$ and a nilpotent shift on the
superdiagonal. When $k = 1$ the block is just the scalar $\lambda$, and the shift
disappears.

## Assembling the form

Applying this basis to every cyclic summand assembles the whole operator.

> **Theorem (Jordan canonical form).** Let $T$ be an operator on a
> finite-dimensional space $V$ over $F$, and assume $F$ contains all eigenvalues of
> $T$. There is a basis of $V$ in which $T$ is block diagonal with a Jordan block
> $J_{k_i}(\lambda_i)$ for each elementary divisor $(x - \lambda_i)^{k_i}$ of
> $V_T$. This matrix is unique up to the order of the blocks along the diagonal.

$$
% caption: The Jordan canonical form: one Jordan block per elementary divisor, stacked along the diagonal, zeros off it.
\begin{tikzpicture}[font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black] (0,0) rectangle (5,5);
\node[draw=acc, fill=acc!12, minimum size=13mm] at (1.0,4.0) {$J_{k_1}$};
\node[draw=acc, fill=acc!12, minimum size=11mm] at (2.5,2.6) {$J_{k_2}$};
\node[draw=acc, fill=acc!12, minimum size=15mm] at (4.1,1.0) {$J_{k_t}$};
\node[black] at (3.7,4.0) {$0$};
\node[black] at (1.0,1.2) {$0$};
\node[black] at (3.4,3.6) {$\ddots$};
\end{tikzpicture}
$$

> **Proof of uniqueness.** The elementary divisors are unique by the structure
> theorem, and permuting the summands only permutes the blocks.[^df-jcf] $\square$
The blocks for a fixed eigenvalue can share that eigenvalue in different sizes, so
the same $\lambda$ may head several blocks.

Because a Jordan block differs from a diagonal matrix only by the superdiagonal
ones, the Jordan form is diagonal exactly when every block has size $1$.

> **Corollary (Diagonalizability).** Assume $F$ contains the eigenvalues of $A$.
> Then $A$ is similar to a diagonal matrix over $F$ if and only if the minimal
> polynomial $m_A(x)$ has no repeated roots. In that case the Jordan form is the
> diagonalization, and it is unique up to permuting the diagonal entries.

> **Proof.** The minimal polynomial is the least common multiple of the blocks'
> polynomials $(x - \lambda)^{k}$, so it is squarefree if and only if every
> $k = 1$, which is when every block is a scalar.[^df-diag] $\square$
Diagonalizability is therefore a property of the minimal polynomial alone, not the
characteristic polynomial: repeated eigenvalues are fine, repeated factors in
$m_A$ are not.

## Generalized eigenspaces

Grouping the Jordan blocks by eigenvalue reassembles the primary decomposition of
$V_T$ in operator language.

> **Definition (Generalized eigenspace).** The **generalized eigenspace** of $T$
> for the eigenvalue $\lambda$ is the direct sum of the cyclic submodules whose
> elementary divisors are powers of $x - \lambda$. It is the $(x-\lambda)$-primary
> component of $V_T$: the set of $v \in V$ annihilated by some power of
> $T - \lambda$.

On the generalized eigenspace, $T - \lambda$ is nilpotent, and its Jordan blocks
are the shift-only matrices with zeros on the diagonal. The kernels of successive
powers of $T - \lambda$ form an increasing flag that stabilizes at the whole
generalized eigenspace.

$$
% caption: On the generalized eigenspace for $\lambda$, the kernels $K_j = \ker(T-\lambda)^j$ nest and grow until they fill the space; the largest Jordan block sets how many steps this takes.
\begin{tikzpicture}[font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[draw=acc, fill=acc!7] (0,0) rectangle (5.4,3.0);
\draw[draw=acc, fill=acc!14] (0,0) rectangle (3.4,3.0);
\draw[draw=acc, fill=acc!22] (0,0) rectangle (1.7,3.0);
\node[acc, font=\footnotesize] at (0.85,1.5) {$K_1$};
\node[acc, font=\footnotesize] at (2.55,1.5) {$K_2$};
\node[black, font=\footnotesize] at (4.4,1.5) {...};
\node[black, anchor=south west, font=\footnotesize] at (0.05,3.1) {generalized eigenspace};
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\node[black, anchor=north, font=\scriptsize] at (3.0,-0.4) {kernels grow with the exponent};
\end{tikzpicture}
$$

The flag makes the block structure computable. Let $r_k = \dim (T - \lambda)^k V$
be the rank of the $k$-th power. The count
$$
r_{k-1} - 2r_k + r_{k+1}
$$
is the number of Jordan blocks of size exactly $k$ for the eigenvalue $\lambda$.
Taking $k = 1$ (with $r_0 = \dim V$) gives the total number of blocks for
$\lambda$, which equals the geometric multiplicity $\dim \ker(T - \lambda)$. This
gives a rank-only route to the Jordan form: compute the ranks of the powers
$(A - \lambda I)^k$ and difference them.[^df-ranks]

## Rational and Jordan forms of the same operator

Both canonical forms come from the same module, through its two decompositions.
Passing between them is the invariant-factor / elementary-divisor conversion of the
first lesson, run on polynomials.

| | Rational canonical form | Jordan canonical form |
| --- | --- | --- |
| Module decomposition | invariant factors $a_1 \mid \cdots \mid a_m$ | elementary divisors $(x-\lambda)^k$ |
| Building block | companion matrix $C_{a_i}$ | Jordan block $J_k(\lambda)$ |
| Field requirement | any field (rational) | must contain the eigenvalues |
| Shape | block companion | nearly diagonal |
| Uniqueness | unique | unique up to block order |
| Diagonal case | rarely diagonal | diagonal iff $m_A$ squarefree |

$$
% caption: The same operator in both canonical forms. The rational form uses companion blocks of the invariant factors; the Jordan form uses blocks of the elementary divisors obtained by splitting those factors over the field.
\begin{tikzpicture}[font=\small, >=stealth]
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\draw[black] (0,0) rectangle (2.6,2.6);
\node[draw=acc, fill=acc!12, minimum size=8mm] at (0.7,1.9) {$C_{a_1}$};
\node[draw=acc, fill=acc!12, minimum size=11mm] at (1.8,0.8) {$C_{a_2}$};
\node[anchor=north, font=\footnotesize, black] at (1.3,-0.1) {rational form};
\draw[->, acc, thick] (3.1,1.3) -- (4.6,1.3) node[midway, above, font=\footnotesize] {split};
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\draw[black] (5.1,0) rectangle (7.7,2.6);
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\node[draw=acc, fill=acc!12, minimum size=7mm] at (6.4,1.3) {$J_{k_2}$};
\node[draw=acc, fill=acc!12, minimum size=7mm] at (7.1,0.6) {$J_{k_3}$};
\node[anchor=north, font=\footnotesize, black] at (6.4,-0.1) {Jordan form};
\end{tikzpicture}
$$

## Worked examples

The three matrices from the
[rational canonical form computation](/abstract-algebra/modules-over-pids/rational-canonical-form)
have eigenvalues $2$ and $3$, both in $\mathbb{Q}$, so their Jordan forms exist
over $\mathbb{Q}$:
$$
A = \begin{pmatrix} 2 & -2 & 14 \\ 0 & 3 & -7 \\ 0 & 0 & 2 \end{pmatrix},
\quad
B = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix},
\quad
C = \begin{pmatrix} 2 & 2 & -1 \\ 0 & 2 & 3 \\ 0 & 0 & 3 \end{pmatrix}.
$$

> **Worked example.** Compute the Jordan forms of $A$, $B$, and $C$.
>
> The invariant factors of $A$ are $x - 2$ and $(x-2)(x-3)$. Splitting the second
> into linear powers gives elementary divisors $x - 2$, $x - 2$, $x - 3$ — all
> first powers, so every Jordan block has size $1$:
>
> $$
> \operatorname{JCF}(A) = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}.
> $$
>
> $A$ is diagonalizable, consistent with its squarefree minimal polynomial
> $(x-2)(x-3)$.
>
> $B$ and $C$ both have the single invariant factor $(x-2)^2(x-3)$, whose
> elementary divisors are $(x-2)^2$ and $x-3$. That yields a size-$2$ Jordan block
> for the eigenvalue $2$ and a size-$1$ block for $3$:
>
> $$
> \operatorname{JCF}(B) = \operatorname{JCF}(C) = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}.
> $$
>
> Their minimal polynomial $(x-2)^2(x-3)$ has a repeated root, so neither is
> diagonalizable, and the shared Jordan form confirms they are similar.

For $3 \times 3$ matrices the characteristic and minimal polynomials determine the
Jordan form. From $4 \times 4$ on they do not, and the rank counts take over.

> **Worked example.** Find the Jordan canonical form of
>
> $$
> D = \begin{pmatrix} 0 & -1 & -1 & 0 \\ 0 & 2 & 0 & 1 \\ 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & 0 \end{pmatrix}
> \quad \text{over } \mathbb{Q}.
> $$
>
> Expanding $\det(xI - D)$ gives the characteristic polynomial $(x-1)^4$, so the
> only eigenvalue is $1$ and the minimal polynomial is $(x-1)^k$ for some
> $k \le 4$. Compute
>
> $$
> D - I = \begin{pmatrix} -1 & -1 & -1 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & -1 \\ 0 & -1 & 0 & -1 \end{pmatrix}
> \ne 0,
> \qquad
> (D - I)^2 = 0,
> $$
>
> so $m_D = (x-1)^2$. Two invariant-factor lists are consistent with
> $c_D = (x-1)^4$ and $m_D = (x-1)^2$:
>
> - $(x-1),\ (x-1),\ (x-1)^2$ — three blocks of sizes $1, 1, 2$;
> - $(x-1)^2,\ (x-1)^2$ — two blocks of sizes $2, 2$.
>
> The rank of $D - I$ decides. In $D - I$, row $4$ is the negative of row $2$, and
> row $3$ is the negative of the sum of rows $1$ and $2$, so only two rows are
> independent: $r_1 = \operatorname{rank}(D - I) = 2$. With $r_0 = 4$ and
> $r_2 = r_3 = 0$, the block counts are
>
> $$
> \underbrace{r_0 - 2r_1 + r_2 = 4 - 4 + 0 = 0}_{\text{blocks of size } 1},
> \qquad
> \underbrace{r_1 - 2r_2 + r_3 = 2 - 0 + 0 = 2}_{\text{blocks of size } 2},
> $$
>
> so the second list is correct: elementary divisors $(x-1)^2, (x-1)^2$ and
>
> $$
> \operatorname{JCF}(D) = \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}.
> $$
>
> The polynomials $c_D$ and $m_D$ alone could not separate the two candidate
> structures; the rank computation supplied the missing invariant.[^df-examples]

> **Worked example.** Count the similarity classes of $4 \times 4$ matrices over
> $\mathbb{Q}$ with characteristic polynomial $(x - 1)^4$. Each class is a
> multiset of elementary divisors $(x-1)^{k_1}, \dots, (x-1)^{k_t}$ with
> $k_1 + \cdots + k_t = 4$ — a partition of $4$. There are five partitions, hence
> five classes:
>
> | partition | Jordan blocks | $m(x)$ | $\operatorname{rank}(A - I)$ |
> | --- | --- | --- | --- |
> | $4$ | $J_4(1)$ | $(x-1)^4$ | $3$ |
> | $3 + 1$ | $J_3(1) \oplus J_1(1)$ | $(x-1)^3$ | $2$ |
> | $2 + 2$ | $J_2(1) \oplus J_2(1)$ | $(x-1)^2$ | $2$ |
> | $2 + 1 + 1$ | $J_2(1) \oplus J_1(1) \oplus J_1(1)$ | $(x-1)^2$ | $1$ |
> | $1 + 1 + 1 + 1$ | $I$ | $x-1$ | $0$ |
>
> The rank column is $4$ minus the number of blocks. The pairs $2+2$ and $2+1+1$
> share the minimal polynomial $(x-1)^2$ and are separated only by the rank — the
> same phenomenon as the matrix $D$ above.

$$
% caption: The five similarity classes with characteristic polynomial $(x - \lambda)^4$: one Jordan structure per partition of $4$. Shaded squares are the Jordan blocks; everything off the blocks is zero.
\begin{tikzpicture}[font=\scriptsize]
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  \draw[acc] (0,0.8) rectangle (0.8,1.6);
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  \node[black, anchor=north] at (0.8,-0.12) {$2 + 2$};
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  \node[black, anchor=north] at (0.8,-0.12) {$2 + 1 + 1$};
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  \node[black, anchor=north] at (0.8,-0.12) {$1 + 1 + 1 + 1$};
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\end{tikzpicture}
$$

## Field dependence

The Jordan form exists only when the eigenvalues lie in $F$, and enlarging the
field can create new blocks. Consider a $6 \times 6$ operator over $\mathbb{Q}$
with an elementary divisor $x^2 + 1$. Over $\mathbb{Q}$ this factor is irreducible,
so it contributes a $2 \times 2$ companion block and no Jordan block exists. Over
$\mathbb{C}$ it splits as $(x - i)(x + i)$, adding two size-$1$ blocks with
eigenvalues $\pm i$. The rational canonical form is the same over both fields; the
Jordan form appears only after adjoining the roots. This is the sense in which the
rational form is "rational" and the Jordan form is not.[^df-field]

When the field is not algebraically closed, the general-purpose canonical form is
the rational one; the Jordan form is the specialization available once the
characteristic polynomial splits.

[^df-jcf]: Dummit & Foote, §12.3 — Theorems 22 and 23: existence of the Jordan canonical form for operators and matrices when $F$ contains the eigenvalues, and uniqueness up to permutation of the blocks.
[^df-diag]: Dummit & Foote, §12.3 — Corollaries 24 and 25: a matrix is diagonalizable if and only if its minimal polynomial has no repeated roots, in which case the Jordan form is diagonal.
[^df-ranks]: Dummit & Foote, §12.3 — Exercises 29 and 30 and the definition of the generalized eigenspace as the $(x-\lambda)$-primary component: the number of Jordan blocks of size $k$ for $\lambda$ is $r_{k-1} - 2r_k + r_{k+1}$ with $r_k = \dim(T - \lambda)^k V$.
[^df-examples]: Dummit & Foote, §12.3 — Examples 1–3: the Jordan forms of the matrices from Example 1 of §12.2, and a $4 \times 4$ matrix with invariant factors $(x-1)^2, (x-1)^2$ illustrating that the characteristic and minimal polynomials do not determine similarity for $n \ge 4$. The matrix $D$ here is a different representative of that same similarity class, chosen so the rank computation is checkable by eye.
[^df-field]: Dummit & Foote, §12.3 — Example 4 and Corollary 18 of §12.2: over $\mathbb{C}$ an elementary divisor $x^2 + 1$ splits into blocks with eigenvalues $\pm i$, while the rational canonical form is unchanged by the field extension.