Why is $A - A^T$ unimodular for any Seifert-matrix $A$ of some knot $K$? I am studying Seifert matrices. Right now I am working with those matrices from a linear algebra perspective, but I will need to understand the knot theory behind it as well. I am using the fact that $A-A^T$ is unimodular for some Seifert-matrix $A$ and I would really appreciate any help on understanding why.
Here is some context:
Let $K$ be some knot in $S^3$, and let $F$ be some Seifert-Surface of $K$. Then as $F$ is orientable we can choose a bicollar $[-1,1]\times F$. The Seifert-Matrix of $K$ according to that Seifert-Surface then is the matrix representing the map
$$f:H_1(F)\times H_1(F)\rightarrow \mathbb{Z}$$
given by
$$f([x],[y])=lk(x,y^+)$$
where lk is the linking form and $y^+$ is the pushoff of $y\times\{0\}$ in the bicollar to $y\times\{1\}$. Now one probable needs to write down a basis (classification of surfaces) and do some calculations, but I am unsure on how to see that the generated matrix has the above property in general.
I am new to knot theory, but rather familiar with a lot of basic algebraic topology. Any help would be greatly appreciated!
Edit: I have already proven that any two generated Seifert-matrices are S-equivalent. Thus it would suffice to show that one generated matrix has the desired property (maybe one of the non-singular ones of minimal size? But that may be nonsense) and then that the property is preserved under S-equivalency. For congruency it is true obviously. I will look into right/left enlargements resp. reductions later. ( This has to be true , else the statement was false)
 A: Consider a genus $g$ oriented surface $S$, with $m$ punctures embedded in $S^3$:

We have a basis for $H_1(S)$ given by the $g$ blue cycles indicated, the $g$ red cycles and the $m$ green cycles.  The intersection form takes a pair of cycles to the number of intersections, counted with sign coming from the orientation of the ordered directions of the cycles, at the point they cross.
From the diagram we see that with respect to the given basis (red, blue green), the intersection form is represented by $Q$: the diagonal composition of $g$ matrices of the form: $$\left(\begin{array}{cc}0&1\\-1&0\end{array}\right),$$
and $m$ $0$'s.
If we quotient by the null space of the form, then we get a unimodular form $\beta\colon\mathbb{Z}^{2g}\times\mathbb{Z}^{2g}\to\mathbb{Z}$.  That is $\beta$ induces an isomorphism: $$\hat{\beta}\colon \mathbb{Z}^{2g}\stackrel{\tiny\sim}\to {\rm Hom}_\mathbb{Z}(\mathbb{Z}^{2g},\mathbb{Z})$$
A matrix $M$ representing $\beta$ (with respect to some basis) also represents the induced map $\hat{\beta}$ (with respect to the dual basis on ${\rm Hom}_\mathbb{Z}(\mathbb{Z}^{2g},\mathbb{Z}))$.
As $\hat{\beta}$ is an isomorphism, with respect to any basis, the matrix representing $\hat{\beta}$ has determinant $\pm 1$.  Thus $M$ has determinant $\pm1$.
It remains to show that $Q=A-A^T$, where $A$ is the Seifert matrix computed with respect to the red\green\blue basis.  Then they will represent the same form, and we know that any matrix representing this form has determinant $\pm1$.
Let $x_i,y_i$ represent the $i$'th red and blue cycles respectively. Thicken the surface $S$, so that $x_i^+,y_i^+$ are the cycles $x_i,y_i$ pushed to the outer surface of $S$.
Then lk$(y_i,x_i^+)-$lk$(x_i,y_i^+)=1$:

The cycles here are not linked about the intersection point.
On the other hand:

The cycles here are linked one degree higher than in the previous diagram, because of the additional link about the intersection point.
If $u,v\in H_1(S)$ do not intersect, then:
$\qquad\qquad\qquad$lk$(u,v^+)=\,$lk$(u,v)=\,$lk$(v,u)=\,$lk$(v,u^+)$.
Thus $A$ is some symmetric matrix plus the diagonal composition of $g$ copies of: $$\left(\begin{array}{cc}0&1\\0&0\end{array}\right),$$
and $m$ $0$'s.
We conclude $Q=A-A^T$ as required.
