Rational Canonical Form and Jordan Form

Key Definitions

Definition / Lemma.
Let $k$ be a field and $A \in M_n(k)$. We may view $k^n$ as a $k[x]$–module by defining

\[f(x) \cdot v := f(A)\, v.\]

We denote this module by $V_A$. Then:

  1. $V_A \cong V_B$ as $k[x]$–modules if and only if $A$ and $B$ are similar matrices.

  2. If $A \in M_n(k)$, $B \in M_m(k)$, and we define the block diagonal matrix

    \[A \oplus B = \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix},\]

    then $V_{A \oplus B} \cong V_A \oplus V_B$.

Proof (Sketch).
A $k[x]$–module isomorphism $\theta : V_A \to V_B$ is also a $k$–linear isomorphism satisfying

\[\theta(x \cdot v) = x \cdot \theta(v), \quad \text{i.e. } \theta(Av) = B\theta(v).\]

Writing $\theta$ as a matrix $P \in \mathrm{GL}_n(k)$ gives $PA = BP$, hence $B = P A P^{-1}$.

Definition / Lemma (Invariant Factors).
Let $A \in M_n(k)$, and let $V_A$ be the corresponding $k[x]$–module. Then $V_A$ decomposes as

\[V_A \cong k[x]/\langle f_1 \rangle \;\oplus \cdots \oplus\; k[x]/\langle f_m \rangle,\]

where $f_1, \ldots, f_m$ are unique monic non-constant polynomials with

\[f_1 \mid f_2 \mid \cdots \mid f_m.\]
  • The $f_i$ are called the invariant factors of $A$.
  • The last one, $f_m$, is the minimal polynomial of $A$. It is the minimal in the sense that if $p(A) = 0$, then $f_m(x) \mid p(x)$.

Proof (Sketch).
Consider the ideal

\[I = \{p(x) \in k[x] : p(A) = 0\}.\]

Since $k[x]$ is a PID, $I = \langle f(x) \rangle$ for some polynomial $f(x)$.
From the decomposition of $V_A$, one checks that $I = \langle f_m(x) \rangle$, hence $f_m$ is the minimal polynomial.

Definition / Lemma (Companion Matrix).
Let $f(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_0 \in k[x]$. The companion matrix of $f$ is

\[C(f) = \begin{bmatrix} 0 & 0 & \cdots & 0 & -c_0 \\ 1 & 0 & \cdots & 0 & -c_1 \\ 0 & 1 & \cdots & 0 & -c_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -c_{n-1} \end{bmatrix}.\]

Then $V_{C(f)} \cong k[x]/\langle f \rangle$.

Proof (Sketch).
Recall that the $k[x]$–module structure on $k^n$ is uniquely determined once we know how $x$ acts.
For $V_A$, this action is defined by $x \cdot v = A v.$

To construct a matrix $C(f)$ such that $V_{C(f)} \cong k[x]/\langle f \rangle$, we want multiplication by $x$ on $k^n$ (i.e. the action of $C(f)$) to mimic the multiplication by $x$ map on $k[x]/\langle f \rangle$. In other words, $C(f)$ must be the matrix representation of the linear map

\[m_x : k[x]/\langle f \rangle \to k[x]/\langle f \rangle, \quad \bar g(x) \mapsto \overline{x g(x)}.\]

To compute this matrix, we pick the natural $k$–basis $\{\bar{1}, \bar{x}, \bar{x}^2, \ldots, \bar{x}^{n-1}\}$ of $k[x]/\langle f \rangle$. Multiplication by $x$ sends these basis elements to $\{\bar{x}, \ \bar{x}^2, \ \ldots, \ \bar{x}^{n-1}, \ \bar{x}^n\}.$ where $\bar x^n \equiv -c_0 - c_1\bar x - \cdots - c_{n-1} \bar x^{n-1}.$ The matrix representation is exactly $C(f)$.

Therefore, multiplication by $x$ on $k[x]/\langle f \rangle$ has exactly the same matrix form as multiplication by $C(f)$ on $k^n$, and so $V_{C(f)} \cong k[x]/\langle f \rangle$ as $k[x]$–modules.

Definition / Theorem (Rational Canonical Form).
Let $A \in M_n(k)$ with invariant factors $f_1, \ldots, f_m$. Then

\[A \sim C(f_1) \oplus C(f_2) \oplus \cdots \oplus C(f_m).\]

This block diagonal form is called the rational canonical form of $A$.

Proof (Sketch).
From the invariant factor decomposition,

\[V_A \cong \bigoplus_{i=1}^m k[x]/\langle f_i \rangle \cong \bigoplus_{i=1}^m V_{C(f_i)} \cong V_{C(f_1) \oplus \cdots \oplus C(f_m)},\]

so $A$ is similar to the block diagonal matrix of companion matrices.

Definition / Theorem (Characteristic Polynomial).
For $A \in M_n(k)$, the characteristic polynomial is

\[f_A(x) = \det(xI - A).\]

If $f_1, \ldots, f_m$ are the invariant factors of $A$, then:

  1. $f_A(x) = \prod_{i=1}^m f_i(x)$.
  2. Every irreducible factor of $f_A(x)$ divides the minimal polynomial $f_m(x)$.
  3. (Cayley–Hamilton Theorem) $f_A(A) = 0$.

Proof (Sketch).
First show that a companion matrix $C(f)$ has characteristic polynomial $f(x)$.
Then, we check that the characteristic polynomial of $A\oplus B$ is $f_A(x)f_B(x)$.

Part 2 and 3 are direct consequences of 1

Definition / Theorem (Jordan Form).
Let $A \in M_n(k)$, and suppose the minimal polynomial splits completely over $k$:

\[f_m(x) = (x - \lambda_1)^{n_{m1}} \cdots (x - \lambda_k)^{n_{mk}}, \quad \lambda_i \in k \text{ distinct}.\]

Then the invariant factors have the form

\[f_i(x) = \prod_{j=1}^k (x - \lambda_j)^{n_{ij}}.\]

By the Chinese Remainder Theorem,

\[k[x]/\langle f_i \rangle \;\cong\; \bigoplus_{j=1}^k k[x]/\langle (x - \lambda_j)^{n_{ij}} \rangle.\]

Hence $V_A$ has the primary decomposition:

\[V_A \;\cong\; \bigoplus_{i=1}^m \bigoplus_{j=1}^k k[x]/\langle (x - \lambda_j)^{n_{ij}} \rangle.\]

We define the Jordan block of size $n$ for eigenvalue $\lambda$ as

\[J_n(\lambda) = \begin{bmatrix} \lambda & 0 & \cdots & 0 \\ 1 & \lambda & \cdots & 0 \\ & \ddots & \ddots & 0\\ & & 1 & \lambda \end{bmatrix}.\]

Then:

  1. $V_{J_n(\lambda)} \cong k[x]/\langle (x - \lambda)^n \rangle$.
  2. $V_A \cong \bigoplus_{i=1}^m \bigoplus_{j=1}^k V_{J_{n_{ij}}(\lambda_j)}$.
  3. $A \sim \bigoplus_{i=1}^m \bigoplus_{j=1}^k J_{n_{ij}}(\lambda_j)$, called the Jordan canonical form of $A$. This form is unique up to reordering the blocks.

Proof (Sketch).
Consider $k[x]/\langle (x - \lambda)^n \rangle$ with basis ${1, (x-\lambda), (x-\lambda)^2, \ldots, (x-\lambda)^{n-1}}$. Multiplication by $x$ has matrix representation $J_n(\lambda)$, giving the desired module isomorphism.