we analyze two particular cases.

-   $p=1$ or $q=1$, and $A(s)\neq 0$ or $B(s)\neq 0$. If ${\mathbb F}\neq \{0, 1\}$, there always exists $P(s)=P_0+sP_1\in {\mathbb F}[s]^{p\times q}$ of $\rank (P(s))=1$ such that $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}P(s)$. For example, if $p=1$, let $c\in {\mathbb F}\setminus \{0\}$ be such that $A(s)\neq cB(s)$. Then $A(s)+(cB(s)-A(s)){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$. If ${\mathbb F}= \{0, 1\}$, then there exists $P(s)\in {\mathbb F}[s]^{p\times q}$ such that $\rank (P(s))=1$ and $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$ if and only if $A(s)\neq B(s)$.

-   $p>1$ or $q>1$, and $A(s), B(s)\in {\mathbb F}[s]^{p \times q}$ are such that $A(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$ and $A(s)\neq 0$. Then there always exists $P(s)\in {\mathbb F}[s]^{p \times q}$ of $\rank( P(s))=1$ such that $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$. For example, let $q=2$ and let $a_1(s)\neq 0$, $a_2(s)$ be the columns of $A(s)$. Then, $B(s){\ensuremath{\stackrel{s.e.}{\sim}}}A(s){\ensuremath{\stackrel{s.e.}{\sim}}}A(s)\begin{bmatrix}1&1\\0&1\end{bmatrix}=
    A(s)+\begin{bmatrix}0& a_1(s)\end{bmatrix}$.

The next lemma shows that in order to solve Problem \[problem\] the pencil $A(s)$ can be substituted by any other pencil strictly equivalent to $A(s)$. It was proven in [@BaRo18 Lemma 3.2] for $p=q$. The proof for the general case is completely analogous.

\[lemmasust\] Let $A(s), B(s), P(s)\in {\mathbb F}[s]^{p\times q}$ be matrix pencils. Let $Q\in \Gl_p({\mathbb F})$, $R\in \Gl_q({\mathbb F})$ and $ A'(s)=Q A(s)R$. If $A(s)+P(s){\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$ then $A'(s)+QP(s)R{\ensuremath{\stackrel{s.e.}{\sim}}}B(s)$.

Problem \[problem\] can be stated as a 