Polynomial clutching families stabilize to linear clutching

For $n=0$, take $L_{0}q=q=0\cdot z+a_0$ and add no summand. Suppose $n\ge 1$. On $(n+1)E$ define the block endomorphism. [F1, construct] $L_{n}q(z)=\left(\begin{array}{ccccc}I& -zI& 0& \cdots & 0\\ 0& I& -zI& \cdots & 0\\ \mathrm{⋮}& & \ddots & \ddots & \mathrm{⋮}\\ 0& \cdots & 0& I& -zI\\ a_n& a_{n-1}& \cdots & a_1& a_0\end{array}\right)\mathrm{.}$ Every entry is a finite polynomial in $z$ and the coefficient bundle maps, and only the superdiagonal entries depend on $z$. Thus $L_{n}q=a(x)z+b(x)$ varies continuously with $x$.

Starting with $L_{n}q$, add $z$ times column $1$ to column $2$, then $z$ times the new column $2$ to column $3$, and continue. The first $n$ rows become the first $n$ rows of the identity, while the final entry of the last row becomes $a_n{z}^n+\cdots +a_{1}z+a_0=q(z)$. Subtract suitable coefficient multiples of the first $n$ rows from the last row to clear its first $n$ entries. The resulting block matrix is $B(z)=\mathrm{diag}(I,\dots ,I,q(z))$.

Each column or row operation in step 2.1 is multiplication by an elementary triangular block matrix. Replacing its off-diagonal entry $c$ by $tc$, $0\le t\le 1$, is a path of invertible elementary matrices. Since $B(z)$ is invertible on $S^1$ by hypothesis, reversing the finite sequence gives a homotopy through invertible clutching maps from $B$ to $L_{n}q$. No fiber bases are selected globally: the block operations are bundle maps, and their invertibility can be checked in any local frame.

By [F1], $B$ clutches $[nE,I]\oplus [E,q]$. By [F2] and step 3.1, the corresponding stabilized bundles satisfy the following isomorphism. [F1, F2, step 3.1] $[E,q]\oplus [nE,I]\cong [(n+1)E,L_{n}q]\mathrm{.}$ This is the promised stable linearization. The matrix homotopy is a finite formula; AC is used only through [F2] to identify the endpoint bundles.