Let be the companion matrix of the irreducible quadratic over a field . Let

where is a identity matrix and is the zero matrix. Here is a **Question** that is alarmingly unobvious: is semisimple?

**Answer:** it depends on the field and on the polynomial. For example, take where is transcendental and let . Then

and an easy calculation shows that . So has irreducible minimal polynomial and so must be semisimple. (Explicitly, if are the basis vectors corresponding to the rows of then the -module decomposes as

.)

Compare with the matrix

.

This matrix has minimal polynomial . One might hope that could be put into Jordan Normal Form (with identity blocks replacing the usual s down the subdiagonal), but of course this is false, because and are not conjugate.

In fact if is any field and is a matrix with minimal polynomial where is irreducible, then can be put into Jordan Normal Form (with identity matrices on the subdiagonal) if and only if 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** does not have a Jordan—Chevalley decomposition as a sum where is semisimple, is nilpotent and and are polynomials in .

**Proof**: if then, since for some we see from the minimal polynomial for that is divisible by . But

and is not semisimple for any .

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 is an irreducible matrix with minimal polynomial . If is separable then one can show by solving a system of equations in and its other derivatives, that there is a polynomial such that if

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

Conversely, suppose that is a cyclic matrix with minimum polynomial . If is not separable then we can write for some polynomial such that and some . If is a nilpotent matrix that is a polynomial in then for some . So if is semisimple then . The left-hand side is equal to , which in turn can be written as plus the product of and a two variable polynomial in and . It follows that is equal to for some polynomial . But has nilpotency degree , and has nilpotency degree at most , so we have a contradiction unless and is semisimple.