= Minimal polynomial of a companion block chain
{title2=$m_B=f^{\lceil N/q\rceil}$}
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 $h'\ne0$; in characteristic zero take $q=1$ and $h=f$. Then
$$
\boxed{m_B(x)=f(x)^{\lceil N/q\rceil},\qquad
\det(xI-B)=f(x)^N.}
$$
Thus 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 $j\geq N$. The <binomial theorem> in positive <field characteristic> gives $B^q=D^q+K^q$. Expanding $h(D^q+K^q)$ gives $f(B)=K^qH$, with $H$ commuting with $K$. The diagonal blocks of $H$ are $h'(A^q)$, 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 $qs\geq N$. 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 $f$.
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