column (row) minimal indices of $T(s)$ are the row (column) minimal indices of $G(s)$.

Observe that if $G(s)\in {\mathbb F}[s]^{p\times q}$ and $\rank (G(s))=p$ ($\rank (G(s))=q$), then $G(s)$ does not have row (column) minimal indices. As a consequence, the invariants for the strict equivalence of regular matrix pencils are reduced to the homogeneous invariant factors.

In this paper we study the Kronecker structure of arbitrary pencils perturbed by pencils of rank one. A matrix pencil of rank one allows a very simple decomposition (see [@GeTr17] for ${\mathbb F}={\mathbb C}$). In the next proposition we analyze this decomposition for arbitrary fields, depending on the Kronecker structure of the pencil.

\[pencil1\] Let $P(s)\in{\mathbb F}[s]^{p\times q}$ be a matrix pencil of $\rank P(s)=1$.

1.  If $P(s)$ has a nontrivial invariant factor, then there exist nonzero vectors $u\in {\mathbb F}^{p}$, $\bar v\in {\mathbb F}^q$ and nonzero pencils $\bar u(s)\in {\mathbb F}[s]^{p}$, $v(s)\in {\mathbb F}[s]^{q}$ such that $$P(s)=uv(s)^T=\bar u(s)\bar v^T.$$

2.  If $P(s)$ has an infinite elementary divisor, then there exist nonzero vectors $u \in {\mathbb F}^{p}$, $v\in {\mathbb F}^q$ such that $$P(s)=uv^T.$$

3.  If $P(s)$ has a positive column minimal index, then there exist a nonzero vector $u\in {\mathbb F}^{p}$ and a nonzero pencil 