Frobenius and Jordan normal form

Let us come back to the situation of a finite-dimensional vector space over a field. Let $k$ be a field, and let $V$ be a finite-dimensional vector space over $k$. Fix a linear map $T : V \to V$. We want to know how $T$ acts on $V$. For instance, can we choose a nice basis of $V$ so that $T$ is represented by a relatively simple matrix?

Here is another way to think about what we are trying to do. First choose an arbitrary basis of $V$, so that $T$ becomes a square matrix. A change of basis is given by a conjugation by an invertible matrix, i.e., $P T P^{-1}$. So we are trying to find an invertible matrix $P$ such that $P T P^{-1}$ is a “relatively simple matrix”.

Our strategy is to combine the data of a vector space $V$ along with the data of a linear map $T : V \to V$, and make it into a data of a single module. Recall the polynomial ring.

Given a vector space $V$ and a linear map $T : V \to V$, we can define a $k[t]$-module structure on $V$ as

$$ (a_0 + \cdots + a_n t^n) \cdot v = a_0 v + a_1 T(v) + a_2 T(T(v)) + \cdots + a_n T^n(v) $$

for $x \in V$ and $a_0, \ldots, a_n \in k$.

Conversely, suppose we are given a $k[t]$-module $V$. Then $V$ is naturally a $k$-vector space because $k \subseteq k[t]$. Also, we can recover the linear map $T : V \to V$ by $T(v) = t \cdot v$. These are inverse constructions, and thus we can translate between these two structures.

$$ \begin{Bmatrix} \text{a } k\text{-vector space } V \\ \text{plus a linear map } T : V \to V \end{Bmatrix} \quad \longleftrightarrow \quad \begin{Bmatrix} \text{a } k[t]\text{-module } V \end{Bmatrix} $$

Because $V$ is a finite-dimensional $k$-vector space, it is finitely generated as a $k[t]$-module. We would like to apply the classification theory of finitely generated modules over a principal ideal domain.

By Theorem 4.3.3, it immediately follows that any $V$(a $k[t]$-module that is a finite-dimensional $k$-vector space) is isomorphic to

$$ V \cong k[t]^{\oplus r} \oplus k[t] / (p_1(t)) \oplus \cdots \oplus k[t] / (p_s(t)) $$

as a $k[t]$-module. But $k[t]$ is an infinite-dimensional $k$-vector space. Because $V$ is finite-dimensional, we must have $r = 0$. Moreover, note that $(c p_i) = (p_i)$ for $c \in k^\times \setminus \lbrace 0\rbrace $ a nonzero constant. This shows that we may assume that $p_i$ are all monic, i.e., have leading coefficient $1$.

What does this structure theory imply? Consider the $i$th component $V_i = k[t] / (p_i(t))$, which is itself a $k[t]$-module and thus corresponds to a vector space plus a linear map. Write

$$ p_i(t) = t^{d_i} + a_{i,d_i - 1} t^{d_i - 1} + \cdots + a_{i,1} t + a_{i,0}. $$

and consider the basis $v_{i,0} = [1]$, $v_{i,1} = [t]$, \dots, $v_{i,d_i-1} = [t^{d_i-1}]$ of $V_i$. Recall that the $k$-linear map $T : V \to V$ is multiplication by $t$. So we have

$$ \begin{aligned} T v_{i,0} &= v_{i,1}, \quad T v_{i,1} = v_{i,2}, \quad \ldots, \quad T v_{i,d_i-2} = v_{i,d_i-1}, \\ T v_{d_i-1} &= [t^{d_i}] = -a_{i,d_i-1} [t^{d_i-1}] - \cdots - a_{i,1} [t] - a_{i,0} [1] \\ &= -a_{i,d_i-1} v_{i,d_i-1} - \cdots - a_{i,1} v_{i,1} - a_{i,0} v_{i,0}. \end{aligned} $$

With this basis $v_{i,0}, \ldots, v_{i,d_i-1}$, the linear map $T : V \to V$ can be represented by the matrix

$$ {}_B[T]^B = \begin{bmatrix} 0 & 0 & \cdots & 0 & -a_{i,0} \\ 1 & 0 & \cdots & 0 & -a_{i,1} \\ 0 & 1 & \cdots & 0 & -a_{i,2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 1 & -a_{i,d_i-1} \end{bmatrix}. $$

Using the classification theorem and applying the above discussion to each component, we obtain the following result.

This also means that any square matrix can be put in a Frobenius normal form by conjugating it with an invertible matrix.

We now look at when $k$ has a particularly nice property.

Here is one example, although we will not prove it.

So if $k$ is algebraically closed, any $k[t]$-module that is a finite-dimensional $k$-vector space looks like

$$ V \cong k[t] / ((t - \alpha_1)^{d_1} \cdots (t - \alpha_k)^{d_k}) \oplus \cdots. $$

At this point, we cannot do much more than the Frobenius normal form. But we are going to use a generalized version of the Chinese remainder theorem to further decompose the module.

Let's apply this theorem to $R = k[t]$.

So if $k$ is algebraically closed, we can write any $k[t]$-module $V$ that is finite-dimensional over $k$ as

$$ V \cong k[t] / ((t - \alpha_1)^{d_1}) \oplus \cdots \oplus k[t] / ((t - \alpha_k)^{d_k}). $$

Here, the numbers $\alpha_i$ need not be distinct from each other. This is because $V$ is a direct sum of several modules that look like $k[t] / (f)$.

Let us now represent the $k[t]$-module $V = k[t] / ((t - \alpha_i)^{d_i})$ in a nice matrix form. Let $T : V \to V$ be multiplication by $t$. We pick the basis

$$ v_{i,1} = [(t - \alpha_i)^{d_i-1}], \quad v_{i,2} = [(t - \alpha_i)^{d_i-2}], \quad \ldots, \quad v_{i,d_i} = [1]. $$

Then the matrix $T$ acts on the vectors as

$$ \begin{aligned} T v_{i,1} &= [t (t - \alpha_i)^{d_i-1}] = [(t - \alpha_i)^{d_i}] + [\alpha_i (t - \alpha_i)^{d_i-1}] = 0 + \alpha_i v_{i,1}, \\ T v_{i,2} &= [t (t - \alpha_i)^{d_i-2}] = [(t - \alpha_i)^{d_i-1}] + [\alpha_i (t - \alpha_i)^{d_i-2}] = v_{i,1} + \alpha_i v_{i,2}, \\ \quad \vdots \\ T v_{i,d_i} &= [t] = [t - \alpha_i] + [\alpha_i] = v_{i,d_i-1} + \alpha_i v_{i,d_i}. \end{aligned} $$

It follows that if we choose the basis $v_{i,1}, \ldots, v_{i,d_i}$, we get the matrix

$$ T = \begin{bmatrix} \alpha_i & 1 & 0 & \cdots & 0 & 0 \\ 0 & \alpha_i & 1 & \cdots & 0 & 0 \\ 0 & 0 & \alpha_i & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & \alpha_i & 1 \\ 0 & 0 & 0 & \cdots & 0 & \alpha_i \end{bmatrix}. $$

From the discussion above, we obtain the following theorem.