Jordan—Chevalley Decomposition

Let $C = \left( \begin{matrix} 0 & 1 \\ -b & -a \end{matrix} \right)$ be the companion matrix of the irreducible quadratic $f(x) = x^2 + ax+b$ over a field $K$. Let

$P = \left( \begin{matrix} C & 0 \\ I & C \end{matrix} \right)$

where $I$ is a $2 \times 2$ identity matrix and $0$ is the zero matrix. Here is a Question that is alarmingly unobvious: is $P$ semisimple?

Answer: it depends on the field and on the polynomial. For example, take $K = \mathbf{F}_2(t)$ where $t$ is transcendental and let $f(x) = x^2 + t$. Then

$P = \left( \begin{matrix} 0 & 1 & 0 & 0 \\ t & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & t & 0 \end{matrix} \right)$

and an easy calculation shows that $f(P) = 0$. So $P$ has irreducible minimal polynomial and so must be semisimple. (Explicitly, if $e_1,e_2,e_3,e_4$ are the basis vectors corresponding to the rows of $P$ then the $K[M]$-module $\langle e_1,e_2,e_3,e_4 \rangle$ decomposes as

$\langle e_1, e_2 \rangle \oplus \langle e_3, e_1+e_4 \rangle$.)

Compare $P$ with the matrix

$Q = \left( \begin{matrix} 0 & 1 & 0 & 0 \\ t & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & t & 0 \end{matrix} \right)$.

This matrix has minimal polynomial $g(x) = (x^2-t)^2$. One might hope that $Q$ could be put into Jordan Normal Form (with identity blocks replacing the usual $1$s down the subdiagonal), but of course this is false, because $P$ and $Q$ are not conjugate.

In fact if $K$ is any field and $P$ is a matrix with minimal polynomial $f(x)^m$ where $f(x) \in K[x]$ is irreducible, then $P$ can be put into Jordan Normal Form (with identity matrices on the subdiagonal) if and only if $f$ is separable. For a nice proof of this see Theorem 6.21 and Corollary 6.22 in Christopher Norman’s recent book Finitely Generated Abelian Groups and Similarity of Matrices over a Field. (I should add that this is one of its harder results. and that the general clarity and level of detail in the book will make it an excellent source for anyone learning the subjects in its title.)

Related to the example above is the following claim.

Claim $Q$ does not have a Jordan—Chevalley decomposition as a sum $Q = S + N$ where $S$ is semisimple, $N$ is nilpotent and $S$ and $N$ are polynomials in $Q$.

Proof: if $N = h(Q)$ then, since $h(Q)^m = 0$ for some $m$ we see from the minimal polynomial for $Q$ that $h(x)$ is divisible by $f(x) = x^2-t$. But

$f(Q) = \left( \begin{matrix} 0 & 0 \\ I & 0 \end{matrix} \right)$

and $Q - rf(Q)$ is not semisimple for any $r \in K$.

In fact the obstruction to the existence of a Jordan—Chevalley decomposition seen in this example is typical of the inseparable case. (This theorem is surely in the literature, but I do not know a reference.)

Theorem A matrix has a Jordan—Chevalley decomposition, in the sense defined above, if and only if it is conjugate to a matrix in Jordan Normal Form.

Outline proof Suppose that $C$ is an irreducible matrix with minimal polynomial $f(x)$. If $f(x)$ is separable then one can show by solving a system of equations in $f'(x)$ and its other derivatives, that there is a polynomial $k(x) \in K[x]$ such that if

$P = \left( \begin{matrix} C & 0 & 0 & \cdots & 0 & 0 \\ I & C & 0 & \cdots & 0 & 0 \\ 0 & I & C & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & C & 0 \\ 0 & 0 & 0 & \cdots & I & C \end{matrix} \right)$

then $k(P)$ is equal to the block matrix with zero matrices on the diagonal and identity matrices below the diagonal. Now take $N = k(P)$ and $S = P - N$. (This extends to a sum of blocks of this form, and hence by primary decomposition to an arbitrary matrix.)

Conversely, suppose that $P$ is a cyclic matrix with minimum polynomial $f(x)^m$. If $f(x)$ is not separable then we can write $f(x) = h(x^p)+r$ for some polynomial $h(x)$ such that $h(0) = 0$ and some $r \in K$. If $N$ is a nilpotent matrix that is a polynomial in $P$ then $N = f(P)q(P)$ for some $q(x) \in K[x]$. So if $S = P - N$ is semisimple then $f(P - f(P)q(P)) = 0$. The left-hand side is equal to $h(P^p - f(P)^pq(P)^p) + r$, which in turn can be written as $f(P)$ plus the product of $f(P)^pq(P)^p$ and a two variable polynomial in $P$ and $f(P)^pq(P)^p$. It follows that $f(P)$ is equal to $f(P)^p a(P)$ for some polynomial $a(X) \in K[x]$. But $f(P)$ has nilpotency degree $m$, and $f(P)^p$ has nilpotency degree at most $m/p$, so we have a contradiction unless $m=1$ and $P$ is semisimple.