= Rational canonical form of an inverse
{title2=$f(t)\mapsto f^\vee(t)=t^df(1/t)/f(0)$}
Let an <invertible matrix> $T$ have monic <invariant factors of a linear operator> $f_j$ of degrees $d_j$. Their constant coefficients are nonzero, since their product is the <characteristic polynomial> and $\det T\ne0$. On a cyclic block, $f(T)=0$ implies $T^{-d}f(T)=0$, so $f^\vee(T^{-1})=0$, where $f^\vee(t)=t^df(1/t)/f(0)$ is monic of degree $d$. The vector $v$ remains cyclic for $T^{-1}$: multiplying $v,T^{-1}v,\ldots,T^{-(d-1)}v$ by the invertible map $T^{d-1}$ gives the reversed original cyclic basis. Thus this polynomial is the <minimal polynomial> of the inverse block.
Normalized reciprocation preserves divisibility because $(fg)^\vee=f^\vee g^\vee$ when both constant coefficients are nonzero. Consequently the <rational canonical form> of $T^{-1}$ is the direct sum of the <companion matrices> $C(f_j^\vee)$. The normalization by $f_j(0)$ is essential to make each reciprocal polynomial monic.
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