$v(s)\in {\mathbb F}[s]^{q}$ such that $$P(s)=uv(s)^T.$$

4.  If $P(s)$ has a positive row minimal index, then there exist a nonzero vector $v\in {\mathbb F}^{q}$ and a nonzero pencil $u(s)\in {\mathbb F}[s]^{p}$ such that $$P(s)=u(s)v^T.$$

[*Proof.*]{} Let $P_c(s)$ be the Kronecker canonical form of $P(s)$. Then, there exist $Q=\begin{bmatrix}q_1&\dots&q_p\end{bmatrix}\in \Gl_{p}({\mathbb F})$ and $R=\begin{bmatrix}r_1^T\\\vdots\\r_q^T\end{bmatrix}\in \Gl_{q}({\mathbb F})$ such that $P(s)=QP_c(s)R$.

1.  If $P(s)$ has a nontrivial invariant factor $s+\lambda$, $\lambda \in {\mathbb F}$, then $P_c(s)=\begin{bmatrix}
    s+\lambda &0 \\0&0
    \end{bmatrix}\in {\mathbb F}[s]^{p\times q}$. Hence, $P(s)=((s+\lambda)q_1)r_1^T=q_1((s+\lambda)r_1)^T.$

2.  If $P(s)$ has an infinite elementary divisor, then $P_c(s)=\begin{bmatrix}
    1&0\\
    0&0\end{bmatrix}\in {\mathbb F}[s]^{p\times q}.$ Therefore, $P(s)=q_1r_1^T$.

3.  If $P(s)$ has a positive column minimal index, then $P_c(s)=
    \begin{bmatrix}
    s&1 &0\\
    0&0 &0\\
    0&0& 0
    \end{bmatrix}\in {\mathbb F}[s]^{p\times q}.$ Therefore, $P(s)=q_1(sr_1^T+r_2^T)$.

4.  If $P(s)$ has a positive row minimal index, then $P(s)^T$ has a positive column minimal index. Therefore, $P(s)^T=q_1(sr_1^T+r_2^T)$, i.e., $P(s)=(sr_1+r_2)q_1^T$, as desired.

$\Box$

Statement of the problem {#secproblem}
========================

The problem we deal with in this paper is the following:

\[problem\] Given two matrix pencils $A(s), B(s)\in {\mathbb F}[s]^{p \times q}$, find necessary and sufficient conditions for the existence of a matrix pencil $P(s)\in {\mathbb F}[s]^{p \times q}$ of $\rank( P(s))=1$ such that $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$.

First of all 