Minimal polynomial of a companion block chain ($m_B=f^{\lceil N/q\rceil }$)

Let $A$ be a companion matrix of a monic irreducible polynomial $f\in F[x]$ of degree $n$. Form the $N$-by-$N$ block matrix $B$ with $A$ on its diagonal, $I_n$ immediately above the diagonal, and zero elsewhere. If the field characteristic is $p>0$, write $f(x)=h(x^q)$ with $q=p^r$ maximal, so that $h^{\prime }\ne 0$; in characteristic zero take $q=1$ and $h=f$. ThenThus a separable polynomial gives $m_B=f^N$, but an irreducible polynomial with zero formal derivative can give a smaller exponent. For example, over the rational function field ${\mathbb{F}}_2(t)$, the irreducible polynomial $f(x)=x^2-t$ has $q=2$, so a two-block chain still has minimal polynomial $f$.

To prove the formula, write $B=D+K$, where $D$ has diagonal blocks $A$ and $K$ is the block shift. These are commuting operators, and $K^j$ vanishes exactly when $j\ge N$. The binomial theorem in positive field characteristic gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$ gives $f(B)=K^{q}H$, with $H$ commuting with $K$. The diagonal blocks of $H$ are $h^{\prime }(A^q)$, which are invertible by Bézout's identity for polynomials, since $h$ is irreducible with nonzero formal derivative and $h(A^q)=0$. Therefore $f(B{)}^s=K^{qs}{H}^s$ vanishes exactly when $qs\ge N$. Finally, the block upper triangular matrix determinant gives the displayed characteristic polynomial, while the Cayley-Hamilton theorem and irreducibility force the minimal polynomial to be a power of $f$.