Seifert stabilization preserves the Levine–Tristram signature

Since $P$ is real, $H_{\omega }(P^{T}VP)=P^T{H}_{\omega }(V)P$. An invertible congruence is a change of coordinates in the associated Hermitian form, so its positive and negative indices are unchanged: a maximal positive subspace maps bijectively by $P$, and likewise for negative subspaces. Hence the signature is invariant under basis change.

Put $c=1-\omega \mathrm{\ne }0$. For either enlargement, the enlarged Hermitian matrix has block form $\tilde{H}=\left(\begin{array}{ccc}H& b& 0\\ b^{\ast }& z(c+\bar{c})& \lambda \\ 0& \bar{\lambda }& 0\end{array}\right),\mathrm{\mid }\lambda \mathrm{\mid }=\mathrm{\mid }c\mathrm{\mid }>0,$ where $H=H_{\omega }(V)$, $b$ is the column determined by the enlargement vectors, and $z\in \mathbb{Z}$. The lower $2\times 2$ block $D$ has determinant $-\mathrm{\mid }c{\mathrm{\mid }}^2<0$, hence one positive and one negative eigenvalue. Its inverse has $(D^{-1}{)}_{11}=0$. The block congruence eliminating $b$ therefore has Schur complement $H-b(D^{-1}{)}_{11}{b}^{\ast }=H$, so $\tilde{H}$ is congruent to $H\oplus D$. The block $D$ contributes signature $0$, proving enlargement invariance for every $\omega \mathrm{\ne }1$, including singular parameters.

By [F1], any two knot matrices are connected by the operations of steps 1.1 and 2.1, so their signatures agree at every nonroot parameter. The one-sided signatures are equal as functions on each punctured neighboring arc; by [F3] their arithmetic averages are defined and agree at a root as well. Reductions reverse the same equalities.