follows that ${\bm{v}} H = 0$ so ${\bm{v}}$ is in the left kernel of $H$. Moreover, restricting ${\bm{v}}$ to its first $n+1$ entries gives $\bar{{\bm{v}}} := (\lambda, \phi_1,\ldots,\phi_n)$, a solution to the simultaneous [Padé]{}problem with $\deg_{- {\bm{N}}} \bar{{\bm{v}}} = \deg_{{\bm{h}}} {\bm{v}}$. Conversely, if $\bar{{\bm{v}}} = (\lambda, \phi_1,\ldots,\phi_n)$ is a solution to the simultaneous [Padé]{}problem, then the extension ${\bm{v}} = (\lambda, \phi_1,\ldots,\phi_n,q_1,\ldots,q_n)$ with $q_i
= (\lambda S_i - \phi_i)/g_i \in {{\mathsf{K}}}[x]$ for $i=1,\ldots,n$ is an approximant of $H$ of order $d$ with $\deg_{{\bm{h}}} {\bm{v}} =
\deg_{-{\bm{N}}} \bar{{\bm{v}}}$.

Finally, consider that a left kernel basis for $H$ is given by $$K = \left[ \begin{array}{c|c} K_1 & K_2 \end{array} \right ] = 
\left [ \begin{array}{cc|c} 1 & {\bm{S}} & \\
 & {  \ifx\null\null  \Gamma_{{\bm{g}}}  \else  \Gamma_{{\bm{g}}_{\null}}  \fi}& -I \end{array} \right ].$$ We must have $G = M K$ for some polynomial matrix $M$ of full row rank. But then $M K_1$ also has full row rank with ${{\textnormal{rowdeg}}}_{-{\bm{N}}} MK_1
= {{\textnormal{rowdeg}}}_{{\bm{h}}} G$.

${\bm{h}} {\leftarrow}-( {\bm{N}} \mid  N_0-1,\ldots,N_0-1) \in {\mathbb Z\xspace}^{2n+1}$ $d {\leftarrow}N_0 + \max_i \deg g_i - 1$ $H = 
  \left[ \begin{array}{c}
       -{\bm{S}} \\
  I \\
  {  \ifx\null\null  \Gamma_{{\bm{g}}}  \else  \Gamma_{{\bm{g}}_{\null}}  \fi}\end{array} \right]$ $(\left [ \begin{array}{c|c} {\bm{\lambda}} & \ast \end{array} 
  \right ], {\bm{\delta}}) 