pencil completion problem, as we see next.

\[lemmaeq\] Let $A(s), B(s)\in {\mathbb F}[s]^{p \times q}$ be matrix pencils such that $A(s)\not {\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$. Then there exists a matrix pencil $P(s)\in {\mathbb F}[s]^{p\times q}$ of $\rank P(s)=1$ such that $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$ if and only if one of the following conditions holds:

1.  There exist matrix pencils $a(s), b(s)\in{\mathbb F}[s]^{1\times q}$, $A_{21}(s)\in {\mathbb F}[s]^{(p-1)\times q}$ such that $ A(s){\ensuremath{\stackrel{s.e.}{\sim}}}\begin{bmatrix}a(s)\\A_{21}(s)\end{bmatrix}$ and $B(s){\ensuremath{\stackrel{s.e.}{\sim}}}\begin{bmatrix}b(s)\\A_{21}(s)\end{bmatrix}$.

2.  There exist matrix pencils $\bar a(s), \bar b(s)\in{\mathbb F}[s]^{p\times 1}$, $A_{12}(s)\in {\mathbb F}[s]^{p\times (q-1)}$ such that $ A(s){\ensuremath{\stackrel{s.e.}{\sim}}}\begin{bmatrix}\bar a(s)&A_{12}(s)\end{bmatrix}$ and $B(s){\ensuremath{\stackrel{s.e.}{\sim}}}\begin{bmatrix}\bar b(s)&A_{12}(s)\end{bmatrix}$.

[*Proof.*]{} Assume that there exists a matrix pencil $P(s)\in {\mathbb F}[s]^{p\times q}$ of $\rank P(s)=1$ such that $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$. By Proposition \[pencil1\], there exist nonzero pencils $ u(s)\in {\mathbb F}[s]^{p}$, $v(s)\in {\mathbb F}[s]^{q}$ such that $P(s)=u(s)v(s)^T$ and $u(s)=u\in {\mathbb F}^{p}$ or $v(s)= v\in {\mathbb F}^{q}$.

If $u(s)=u\in {\mathbb F}^{p}$, let $R\in \Gl(p)$ be such that $Ru=\begin{bmatrix}1\\0\end{bmatrix}\in{\mathbb F}^{(1+(p-1))}$ and let $RA(s)=\begin{bmatrix}a(s)\\A_{21}(s)\end{bmatrix}\in {\mathbb F}[s]^{(1+(p-1))\times q}$ and $b(s)=a(s)+v(s)^T$. Then $A(s){\ensuremath{\stackrel{s.e.}{\sim}}}\begin{bmatrix}a(s)\\A_{21}(s)\end{bmatrix}$ and $B(s){\ensuremath{\stackrel{s.e.}{\sim}}}R(A(s)+P(s))=\begin{bmatrix}a(s)\\A_{21}(s)\end{bmatrix}+\begin{bmatrix}v(s)^T\\0\end{bmatrix}=\begin{bmatrix}b(s)\\A_{21}(s)\end{bmatrix}$. Therefore, (i) holds.

If $v(s)= v\in {\mathbb F}^{q}$, we can analogously obtain (ii).

Conversely, let us assume that (i) holds. As $A(s)\not {\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$, we have $a(s)\neq b(s)$. Let $\bar P(s)=\begin{bmatrix}b(s)-a(s)\\0\end{bmatrix}\in {\mathbb F}[s]^{(1+(p-1))\times q}$. Then $\rank \bar P(s)=1$ and $\begin{bmatrix}a(s)\\A_{21}(s)\end{bmatrix}+\bar P(s)= \begin{bmatrix}b(s)\\A_{21}(s)\end{bmatrix}$. By 