One crossing change changes a Seifert matrix by rank at most one

Switching over and under does not change the oriented smoothing at the distinguished crossing. Thus [F1] gives the same Seifert circles, disk records, and bands at every other crossing. Only the distinguished band changes: its two possible half-twist directions differ by one full twist. By [F2], both disk-and-band constructions give connected embedded oriented spanning surfaces, and each surface retracts to the finite graph with one vertex per disk and one edge per band.

Let $e$ be the distinguished edge of that graph. If $e$ is a bridge, every graph cycle avoids $e$, so choose the corresponding homology basis outside the changed band; all Seifert-pairing entries are then identical. If $e$ is not a bridge, the graph with $e$ removed is connected. Choose a spanning tree of that graph; its fundamental cycles, together with the fundamental cycle using $e$, form an integral homology basis. All basis curves except the last can be represented disjoint from the distinguished band.

The crossing change is supported in a ball disjoint from every other band and disk. Hence the basis curves avoiding $e$, and their positive push-offs, have identical linking numbers before and after the change. Their pairings with the last basis curve also remain the same, because the other curve and its push-off miss that ball. The last curve passes once through the distinguished band; changing the band by a full twist adds one signed meridional turn to its positive push-off about its core, so its self-linking in [F3] changes by exactly $+1$ or $-1$. Thus only one diagonal entry changes. The matrix difference is zero in the bridge case and is $\pm v{v}^T$ in the non-bridge case, so its rank is at most one.