Rational Canonical Form I - Definition

The structure of $F [x]$ -modules

Suppose $F$ is a field and $V$ is a finite-dimensional $F$ -vector space. Let $T : V \to V$ be a fixed linear endomorphism. We can then consider $V$ as an $F [x]$ -module, with the action of $x$ on $V$ given by the linear endomorphism $T$ . In other words, for any vector $v \in V$ we have $x \cdot v = T (v)$ . The general polynomial $p (x) = a_{0} + a_{1} x + \dots + a_{n} x^{n} \in F [x]$ then acts by

p (x) \cdot v = a_{0} v + a_{1} T (v) + \dots + a_{n} T^{n} (v),

where $T^{k} = T \circ \dots \circ T$ is repeated composition.

Since $V$ is finitely generated as an $F$ -module, it is also finitely generated as an $F [x]$ -module. The fundamental structure theorem for modules over a PID then provides a direct sum decomposition of $V$ as an $F [x]$ -module. The free part of this decomposition must be trivial, since every nonzero free $F [x]$ -module is infinite-dimensional as an $F$ -vector space (since $F [x]$ is infinite-dimensional as an $F$ -vector space). So, $V$ must be isomorphic to a direct sum of cyclic torsion $F [x]$ -modules

V ≃ F [x] / ⟨ a_{1} (x) ⟩ \oplus \dots \oplus F [x] / ⟨ a_{m} (x) ⟩,

where $a_{1} (x), \dots, a_{m} (x) \in F [x]$ are nonzero nonunits (i.e., nonconstant polynomials) and $a_{1} (x) ∣ a_{2} (x) ∣ \dots ∣ a_{m} (x)$ . Recall that the $a_{i} (x)$ are called invariant factors of the $F [x]$ -module. These invariant factors are unique up to unit (which are the constant polynomials); if we require them to be monic, then they are unique.

Note that the annihilator of $V$ (as a torsion $F [x]$ -module) is the ideal $⟨ a_{m} (x) ⟩ \subseteq F [x]$ . Note that this means $a_{m} (x) \cdot v = 0_{V}$ for every $v \in V$ . By the definition of our action, this means that the endomorphism $a_{m} (T)$ is identically zero on the entire vector space $V$ . Since we've currently assuming $a_{m} (x)$ is monic, this is exactly the minimal polynomial of $T$ .

The minimal polynomial of

T

The minimal polynomial $m_{T} (x)$ of the linear endomorphism $T$ is $a_{m} (x)$ , the largest invariant factor in the above decomposition.

A warning about bases and generators

Suppose $B = {v_{1}, \dots, v_{k}}$ is a basis for $V$ as an $F$ -vector space. Then the set $B$ generates $V$ as an $F$ -vector space, and so every $v \in V$ can be written as an $R$ -linear combination of the vectors in $B$ . It follows that every $v \in V$ can also be written as an $F [x]$ -linear combination of the vectors in $B$ ; i.e., the same set $B$ generates $V$ as an $F [x]$ -module.

However, the set $B$ is not $F [x]$ -linearly independent. In fact, as an $F [x]$ -module the space $V$ is torsion and so there are no nonempty $F [x]$ -linearly independent sets in $V$ ! Consequently, as an $F [x]$ -module the space $V$ doesn't have a basis.

The rational canonical form

Consider one of the direct summands $F [x] / ⟨ a (x) ⟩$ , where $a (x) = b_{0} + b_{1} x + \dots + b_{k - 1} x^{k - 1} + x^{k}$ . As an $F$ -vector space, a basis for $F [x] / ⟨ a (x) ⟩$ is the set ${1, \overset{―}{x}, \dots, {\overset{―}{x}}^{k - 1}}$ , where $\overset{―}{x} = x + ⟨ a (x) ⟩$ is shorthand notation for the coset represented by $x$ . Under this basis, the action of $x$ is simply:

\begin{aligned} 1 & \mapsto \overset{―}{x} \\ \overset{―}{x} & \mapsto {\overset{―}{x}}^{2} \\ {\overset{―}{x}}^{2} & \mapsto {\overset{―}{x}}^{3} \\ ⋮ \\ {\overset{―}{x}}^{k - 2} & \mapsto {\overset{―}{x}}^{k - 1} \\ {\overset{―}{x}}^{k - 1} & \mapsto {\overset{―}{x}}^{k} = - b_{0} - b_{1} \overset{―}{x} - \dots - b_{k - 1} {\overset{―}{x}}^{k - 1} . \end{aligned}

So, with respect to our chosen basis, the matrix for the action of $x$ is

[\begin{matrix} 0 & 0 & \dots & \dots & - b_{0} \\ 1 & 0 & \dots & \dots & - b_{1} \\ 0 & 1 & \dots & \dots & - b_{2} \\ ⋮ & ⋮ & ⋱ & \dots & ⋮ \\ 0 & 0 & \dots & 1 & - b_{k - 1} \end{matrix}]

This matrix is called the companion matrix of $a (x)$ and is denoted $C_{a (x)}$ .

If we repeat this process for every direct summand in the invariant factor decomposition of $V$ , we see that the matrix for $T$ with respect to that choice of basis (amalgamated from the bases for each summand, as described above) is

[\begin{matrix} C_{a_{1} (x)} \\ C_{a_{2} (x)} \\ ⋱ \\ C_{a_{m} (x)} \end{matrix}]

Definition of rational canonical form

A matrix is said to be in rational canonical form if it is the direct sum of companion matrices for nonconstant monic polynomials $a_{1} (x), \dots, a_{m} (x)$ with $a_{1} (x) ∣ a_{2} (x) ∣ \dots ∣ a_{m} (x)$ . These polynomials are called the invariant factors of the matrix.

A rational canonical form for a linear endomorphism $T$ is a matrix representing $T$ that is in rational canonical form.

Example

Suppose $V$ is a finite-dimensional $Q$ -vector space and $T : V \to V$ is a $Q$ -linear transformation. Then $V$ can be given the structure of a $Q [x]$ -module by letting $x$ act via $T$ . Suppose the invariant factors of $V$ as a $Q [x]$ -module are

\begin{aligned} a_{1} (x) & = 3 + x, \\ a_{2} (x) & = (3 + x) (2 + x) = 6 + 5 x + x^{2}, \\ a_{3} (x) & = (3 + x)^{2} (2 + x) = 18 + 21 x + 8 x^{2} + x^{3} . \end{aligned}

In other words, there is a $Q [x]$ -module isomorphism

ϕ : V \tilde{\to} (Q [x] / ⟨ 3 + x ⟩) \oplus (Q [x] / ⟨ 6 + 5 x + x^{2} ⟩) \oplus (Q [x] / ⟨ 18 + 21 x + 8 x^{2} + x^{3} ⟩) .

This is an isomorphism of $Q [x]$ -modules, where the action of $x$ on the left is via $T$ and the action of $x$ in each summand on the right is by multiplication by $\overset{―}{x}$ , the coset represented by $x$ in the given quotient ring.

Let's look at each summand in turn. In the quotient $Q [x] / ⟨ 3 + x ⟩$ we have $3 + \overset{―}{x} = 0$ . So, if we use the $Q$ -basis $B_{1} = {1}$ then the action of $x$ is given by $1 \mapsto \overset{―}{x} = - 3$ . The matrix for the action of $x$ on this summand is therefore the $1 \times 1$ matrix

C_{a_{1} (x)} = [\begin{matrix} - 3 \end{matrix}] .

In the summand $Q [x] / ⟨ 6 + 5 x + x^{2} ⟩$ we have $6 + 5 \overset{―}{x} + {\overset{―}{x}}^{2} = 0$ . So, if we use the $Q$ -basis $B_{2} = {1, \overset{―}{x}}$ then the action of $x$ is given by $1 \mapsto \overset{―}{x}$ and $\overset{―}{x} \mapsto {\overset{―}{x}}^{2} = - 6 - 5 \overset{―}{x}$ . The matrix for the action of $x$ on this summand is therefore the $2 \times 2$ matrix

C_{a_{2} (x)} = [\begin{matrix} 0 & - 6 \\ 1 & - 5 \end{matrix}] .

In the summand $Q [x] / ⟨ 18 + 21 x + 8 x^{2} + x^{3} ⟩$ we have $18 + 21 \overset{―}{x} + 8 {\overset{―}{x}}^{2} + {\overset{―}{x}}^{3} = 0$ . So, if we use the basis $B_{3} = {1, \overset{―}{x}, {\overset{―}{x}}^{2}}$ then the action of $x$ is given by

\begin{aligned} 1 & \mapsto \overset{―}{x} \\ \overset{―}{x} & \mapsto {\overset{―}{x}}^{2} \\ {\overset{―}{x}}^{2} & \mapsto {\overset{―}{x}}^{3} = - 18 - 21 \overset{―}{x} - 8 {\overset{―}{x}}^{2} . \end{aligned}

The matrix for the action of $x$ on this summand is therefore the $3 \times 3$ matrix

C_{a_{3} (x)} = [\begin{matrix} 0 & 0 & - 18 \\ 1 & 0 & - 21 \\ 0 & 1 & - 8 \end{matrix}] .

With all of this in mind, the isomorphism $ϕ$ corresponds to a basis $B = {v_{1}, v_{2}, v_{3}, v_{4}, v_{5}, v_{6}}$ for $V$ as an $F$ -vector space such that the matrix for $T$ with respect to this basis $B$ is

M (T; B) = [\begin{matrix} - 3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & - 6 & 0 & 0 & 0 \\ 0 & 1 & - 5 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & - 18 \\ 0 & 0 & 0 & 1 & 0 & - 21 \\ 0 & 0 & 0 & 0 & 1 & - 8 \end{matrix}] .

Also note that the $Q$ -vector space $V$ decomposes into a direct sum of three subspaces, namely $V_{1} = span {v_{1}}$ , $V_{2} = span {v_{2}, v_{3}}$ , and $V_{3} = span {v_{4}, v_{5}, v_{6}}$ . These three spaces are invariant under the action of the linear transformation $T .$ Indeed, we have

\begin{aligned} T (v_{1}) & = - 3 v_{1}, \\ T (v_{2}) & = v_{3} and T (v_{3}) = - 6 v_{2} - 5 v_{3} \\ T (v_{4}) & = v_{5} and T (v_{5}) = v_{6} and T (v_{6}) = - 18 v_{4} - 21 v_{5} - 8 v_{6} . \end{aligned}

Suggested next note

Rational Canonical Form II - Additional Properties