The first Frobenius companion pencil preserves the finite and infinite polynomial spectrum

For $m\ge 2$, partition the columns of $L$ as $u=(z_1,\dots ,z_{m-1})$ and $x=z_m$, and the rows as the top block row and the lower $m-1$ block rows; write the resulting blocks as $L=\left[\begin{array}{cc}C& E\\ D& G\end{array}\right]$. The lower-left block $D(\lambda )$ is upper block-bidiagonal with $-I_n$ on its diagonal and $\lambda I_n$ on its superdiagonal. Thus $D=-(I-\lambda S)$ for the nilpotent upper block shift $S$, so $D^{-1}=-(I+\lambda S+\cdots +{\lambda }^{m-2}{S}^{m-2})$ is polynomial and $\det D=(-1{)}^{n(m-1)}$. The lower equations with zero right-hand side are $Du+Gx=0$; solving them gives $z_j={\lambda }^{m-j}x$. Substitution into the top row shows that the Schur complement is $E-C{D}^{-1}G=P(\lambda )$. To compute the determinant, swap the top and lower row groups, with sign $(-1{)}^{n^2(m-1)}$, obtaining $\left[\begin{array}{cc}D& G\\ C& E\end{array}\right]$. Left multiplication by $\left[\begin{array}{cc}I& 0\\ -C{D}^{-1}& I\end{array}\right]$ has determinant one and gives $\left[\begin{array}{cc}D& G\\ 0& P\end{array}\right]$, whose determinant is $\det D\det P$. Since $\det D=(-1{)}^{n(m-1)}$, the total sign is $(-1{)}^{n^2(m-1)+n(m-1)}=(-1{)}^{n(n+1)(m-1)}=1$, so $\det L=\det P$. For $m=1$, $L=P$ directly.

For the partition in step 1.1, left multiplication by $\left[\begin{array}{cc}I& -C{D}^{-1}\\ 0& I\end{array}\right]$ clears the top-left block and leaves $P$ in the top-right block, since $E-C{D}^{-1}G=P$. Left multiplication of the lower rows by $D^{-1}$ makes the lower-left block $I$ and the lower-right block $H:=D^{-1}G$. Right multiplication by $\left[\begin{array}{cc}I& -H\\ 0& I\end{array}\right]$ clears that lower-right block; a block-column permutation then gives $\mathrm{diag}(P,I_{(m-1)n})$. Each operation and its inverse has polynomial entries and determinant a nonzero constant, so their products are unimodular.

Put $L_h(\alpha ,\beta ):=\beta A_c-\alpha B_c$. If $\beta \mathrm{\ne }0$, then $L_h=\beta L(\alpha \mathrm{/}\beta )$, so step 1.1 gives $\det L_h={\beta }^{mn}\det P(\alpha \mathrm{/}\beta )=\det P_h(\alpha ,\beta )$. Both sides are polynomials in $(\alpha ,\beta )$. Fix any $\beta \mathrm{\ne }0$; the identity holds for every $\alpha$, so their coefficients in powers of $\alpha$ agree. Each such coefficient is a polynomial in $\beta$ and agrees for every $\beta \mathrm{\ne }0$; as univariate polynomials with infinitely many common values they are identical, so the identity holds also at $\beta =0$. For finite $\lambda$, the lower rows of $L(\lambda )z=0$ force $z_j={\lambda }^{m-j}{z}_m$; if $z\mathrm{\ne }0$ then $z_m\mathrm{\ne }0$, and the top row is $P(\lambda )z_m=0$. Conversely, for each nonzero $x$ with $P(\lambda )x=0$, the lift in [F2] is nonzero and the lower rows vanish while the top row is $P(\lambda )x$. At infinity, $L_h(1,0)=-B_c=\mathrm{diag}(A_m,I_n,\dots ,I_n)$, so its kernel consists exactly of $(x,0,\dots ,0{)}^T$ with $A_{m}x=0$.

For the reversal pencil $\mu A_c-B_c$, reorder the block columns as $(z_2,\dots ,z_m,z_1)$. Its lower equations are $z_i-\mu z_{i-1}=0$; on the first $m-1$ reordered block columns their coefficient matrix is lower block-bidiagonal with $I_n$ on its diagonal and $-\mu I_n$ on its subdiagonal, so its determinant is $1$ and its inverse is polynomial by the finite nilpotent-shift expansion. Solving these equations gives $z_i={\mu }^{i-1}{z}_1$; substitution into the top row gives ${\sum }_{j=0}^m{A}_{m-j}{\mu }^j=P_m^{\mathrm{r}\mathrm{e}\mathrm{v}}(\mu )$. The same row elimination, normalization by this polynomial inverse, column clearing, and block permutation used in step 2.1 therefore give $\mathrm{diag}(P_m^{\mathrm{r}\mathrm{e}\mathrm{v}},I_{(m-1)n})$ by unimodular polynomial matrices. For $m=1$ the reversal pencil equals $P_1^{\mathrm{r}\mathrm{e}\mathrm{v}}$ directly.

The reversal determinant identity is $\det P_m^{\mathrm{r}\mathrm{e}\mathrm{v}}(\mu )={\mu }^{mn}\det P(1\mathrm{/}\mu )$. If $d=\mathrm{deg}\det P$, write $\det P(\lambda )=c_d{\lambda }^d+\cdots +c_0$ with $c_d\mathrm{\ne }0$; then $\det P_m^{\mathrm{r}\mathrm{e}\mathrm{v}}(\mu )=c_d{\mu }^{mn-d}+\cdots +c_0{\mu }^{mn}$, whose zero order at $0$ is exactly $mn-d$. Step 3.1 shows the reversal pencil has the same zero order, and step 2.2 identifies the homogeneous companion spectrum and eigenspaces. In the permutation expansion of $\det P(\lambda )$, every product has $n$ factors of degree at most $m$; degree $mn$ can occur only by taking the ${\lambda }^m$ coefficient from each factor, so its coefficient is $\det A_m$. Thus invertibility of $A_m$ is equivalent to $d=mn$ and no infinity eigenvalue; singularity gives $d<mn$ and positive multiplicity at infinity. The geometric multiplicity is $\dim \ker A_m$ by step 2.2 and is not used to infer the zero order.