The genus $g(K)$ of a knot $K$ is the minimum possible (topological) genus $g(S) = \frac{2-\chi(S)-B}{2}$ of a Seifert surface $S$ for the knot $K$ (where $\chi(S)$ denotes the Euler characteristic of $S$ and $B$ is the number of bands of $S$).
The Alexander polynomial $\Delta_K(t)$ of a knot $K$ is defined as $\Delta_K(t) = \det(A_K(t))$ where $A_K(t)$ is an Alexander matrix of the knot $K$.
In short, I desire to show $\deg(\Delta_K) \leq 2 g(K)$.
I know that $\Delta_K(-1) = \det(K)$ yields the determinant of the knot $K$, another invariant. I know that the alexander polynomial can also be seen as $\Delta_K(t) = \det(V-tV^T)$ where $V$ is a Seifert matrix for $S$. However, I don't quite see how to start a reasoning for proving my claim. I am not in search of solutions, but any help that helps me get some oversight in how this may be achieved is highly apprechiated. (Can the signature of a knot be used maybe, since it also is bounded by $g(K)$?)
The Leibniz formula for determinants implies that the map $t \mapsto \det(A-tB)$ for square $n \times n$ matrices $A,B$ is a polynomial of at most degree $n$ (this degree can be achieved, e.g. by $\det(\mathbf{0}-t(-\mathbf{1}))=\det(t\mathbf{1})=t^n\det(\mathbf{1})=t^n$).
The Seifert matrix $V$ (and thus its symmetrization $V+V^T$, too) is a matrix of size $2g \times 2g$ where $g=g(K)$ is the genus of the knot. By construction, $V+V^T$ can have at most $2g$ distinct eigenvalues, proving a bound on the knot signature: $|\sigma(K)| \leq 2g(K)$ (as a nice side result). In addition, the beforementioned fact proves the assertion when setting $A=V$, $B=V^T$.