Linear algebra | Lovely little lemmas

I needed the following well-known result in the course I’m teaching:

Lemma. Let $k$ be a field and $M \in M_n(k)$ a nilpotent matrix. Then $\operatorname{tr} M = 0$ .

The classical proof uses the flag of subspaces

$0 \subseteq \ker M \subseteq \ker M^2 \subseteq \ldots \subseteq \ker M^n = k^n$

to produce a basis for which $M$ is upper triangular. Here is a slick basis-independent commutative algebra proof that shows something better:

Lemma. Let $R$ be a commutative ring with nilradical $\mathfrak r$ , and let $M \in M_n(k)$ be a nilpotent matrix. Then the characteristic polynomial $p_M(t) = \det(tI-M)$ satisfies

$p_M(t) \in t^n + \mathfrak r[t]_{< n}.$

Here we write $I[t]_{<n} \subseteq R$ for the polynomials in $R$ of degree smaller than $n$ whose coefficients lie in a given ideal $I \subseteq R$ .

Note that the formulation should ring a bell: in the previous post we saw that $R[t]^\times = R^\times + t\mathfrak r[t]$ . When $R$ is a domain, this reduces to $R[t]^\times = R^\times$ , and the lemma just says that $p_M(t) = t^n$ .

This suggests that we shouldn’t work with $\det(tI-M)$ but with its anadrome (or reciprocal) $t^n\det(t^{-1}I-M) = \det(I-tM)$ .

Proof of Lemma. We have to show that $\det(I-tM) \in 1 + t\mathfrak r[t]$ . Since $M$ is nilpotent, there exists $r \in \mathbf N$ with $M^{r+1} = 0$ , so $(I-tM)(I+tM+\ldots+t^rM^r) = I$ . Thus $\det(I-tM) \in R[t]^\times = R^\times + t\mathfrak r[t]$ . Evaluating at $t=0$ shows that the constant coefficient is 1. $\qedsymbol$

Lovely little lemmas

Or maybe ‘amusing observations’

Category Archives: Linear algebra

Traces of nilpotent matrices