above computation the row space of $\hat A{^\top}$ must be a subset of the row space of $G$. But since $G \hat B = (x^dI_m) R$ for some $R \in {{\mathsf{K}}}[x]^{(n+1)\times(n+1)}$, then $\det G = x^d \det R$. Thus $x^d \mid \det G$. But $\det \hat A{^\top}= x^d$, so the row space of $\hat A{^\top}$ can not be smaller than the row space of $G$. That is, $\hat A{^\top}$ is an approximant basis for $B$ of order $d$. Taking the transpose through the argument shows that $\hat
B{^\top}$ is an approximant basis of $\hat B$ of order $d$.

\[thm:dualityMinbasis\] Let $A$ and $B$ be as follows. $$\begin{aligned}
    A &= \left [ \begin{array}{cccc}
                   -{\bm{S}} \\{  \hline  \\[\dimexpr-\normalbaselineskip+1pt]  }
                   I
                 \end{array} \right ] \in {{\mathsf{K}}}[x]^{(n+1) \times (n+1)}
    &&&
    \hspace*{-1em}B  &= \left [ \begin{array}{c}
                    1 \\ {\bm{S}}
                  \end{array} \right] \in {{\mathsf{K}}}[x]^{(n+1) \times 1}\end{aligned}$$ If $G$ is an ${\bm{N}}$-minimal approximant basis of $B$ of order $d$ with shift ${\bm{N}} \in {\mathbb Z\xspace}_{\geq 0}^{n+1}$, then ${{\textnormal{adj}}}(G{^\top})$ is a $(-{\bm{N}})$-minimal approximant basis of $A$ of order $d$. Moreover, if ${\bm{\eta}} = {{\textnormal{rowdeg}}}_{{\bm{N}}} G$, then ${{\textnormal{rowdeg}}}_{-{\bm{N}}} {{\textnormal{adj}}}(G) =(\eta - N - \eta_1,\ldots , \eta - N
-\eta_{n+1})$, where $\eta = \sum_i \eta_i$ and $N = \sum_i N_i$.

