A companion-pencil residual transfers to the polynomial residual through explicit block factors

From the lower residual equations in [F2], $z_{k-1}=\lambda z_k-e_k$. Back-substitution from $x=z_m$ gives $z_j={\lambda }^{m-j}x-{\sum }_{k=j+1}^m{\lambda }^{k-j-1}{e}_k$ for $1\le j<m$; when $j=m$ the same formula has an empty sum.

Substitute these expressions into the top row from [F2]. The terms containing $x$ sum to ${\sum }_{j=0}^m{A}_j{\lambda }^{j}x=P(\lambda )x$. For fixed $e_k$, its coefficient is the negative of ${\sum }_{j=m-k+1}^m{A}_j{\lambda }^{j-m+k-1}$: the $z_1$ term contributes $-(\lambda A_m+A_{m-1}){\lambda }^{k-2}{e}_k$, and each $z_j$ term for $2\le j<k$ contributes $-A_{m-j}{\lambda }^{k-j-1}{e}_k$. Thus $e_1=P(\lambda )x-{\sum }_{k=2}^m{Q}_k(\lambda )e_k$, which is the claimed identity.

We first derive the finite-dimensional Euclidean estimates used here. For $u,v\in {\mathbb{C}}^d$ with $v\mathrm{\ne }0$, set $c:=⟨u,v⟩\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$. Direct expansion gives $\mathrm{\parallel }u-cv{\mathrm{\parallel }}_2^2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2-\mathrm{\mid }⟨u,v⟩{\mathrm{\mid }}^2\mathrm{/}\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\ge 0$, hence $\mathrm{\mid }⟨u,v⟩\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2$; if $v=0$ the inequality is immediate. Since $\mathrm{\parallel }u+v{\mathrm{\parallel }}_2^2=\mathrm{\parallel }u{\mathrm{\parallel }}_2^2+\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+2\mathrm{Re}⟨u,v⟩\le (\mathrm{\parallel }u{\mathrm{\parallel }}_2+\mathrm{\parallel }v{\mathrm{\parallel }}_2{)}^2$, taking nonnegative square roots gives $\mathrm{\parallel }u+v{\mathrm{\parallel }}_2\le \mathrm{\parallel }u{\mathrm{\parallel }}_2+\mathrm{\parallel }v{\mathrm{\parallel }}_2$. For any matrix $M$ and vector $v$, apply the derived inner-product bound to each row of $M$ and $\bar{v}$ to obtain $\mathrm{\mid }(Mv{)}_r{\mathrm{\mid }}^2\le ({\sum }_s\mathrm{\mid }{M}_{rs}{\mathrm{\mid }}^2)\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$; summing over rows gives $\mathrm{\parallel }Mv{\mathrm{\parallel }}_2\le \mathrm{\parallel }M{\mathrm{\parallel }}_F\mathrm{\parallel }v{\mathrm{\parallel }}_2$. Now put $y=e_1+{\sum }_{k=2}^m{Q}_k{e}_k=P(\lambda )x$. Repeated use of the derived triangle inequality gives $\mathrm{\parallel }y{\mathrm{\parallel }}_2\le \mathrm{\parallel }{e}_1{\mathrm{\parallel }}_2+{\sum }_{k=2}^m\mathrm{\parallel }{Q}_k{e}_k{\mathrm{\parallel }}_2$. Applying the rowwise matrix estimate to each $Q_k$ gives the claimed bound. For the companion lift, each lower row vanishes by construction and the top row is $P(\lambda )x$.

To see why no uniform unscaled transfer factor follows, take the fixed scalar polynomial $P(\lambda )={\lambda }^2+1$. For any $t>0$, set $x=t\mathrm{/}(t^2+1)$, $z_2=x$, and $z_1=tx-1$. Direct substitution into its first companion pencil gives $e_1=t{z}_1+z_2=0$ and $e_2=-z_1+t{z}_2=1$, while $P(t)x=t$. Hence the block-pencil residual norm is $1$ and the polynomial residual is $t$, which is unbounded as $t\to \mathrm{\infty }$. Any error or conditioning comparison must retain the displayed coefficient and parameter scaling.