Can we recover the signature of a knot from an Alexander module presentation matrix? Let $K$ be a knot in $S^3$ and let $A(t)$ be a presentation matrix for the Alexander module of $K$.  Is there a way to use $A(t)$ to find the signature of $K$?
 A: Having a presentation matrix for the Alexander module is not enough to determine the signature. For instance, both the left and right handed trefoil knots admit $t-1+t^{-1}$ as presentation matrices for their the Alexander modules, but their signatures differ.
Put differently, a presentation matrix certainly determines the Alexander module, however this latter module does not determine the signature. However the pair consisting of the module and the Blanchfield pairing does determine the signature. 
In fact, the whole Levine-Tristram signature function can be recovered from a any Hermitian non-degenerate matrix A(t) that presents the Blanchfield pairing (not just the Alexander module) via the formula 
$$ \sigma_K(\omega)=\sigma(A(\omega))-\sigma(A(1)).$$
The above mirror image trick does not work anymore: two Hermitian non-degenerate matrices A(t) and -A(t) will present the same module, but not the same linking form. Note also that a presentation matrix A(t) need not arise from a Seifert matrix via the formula tA-A^T!
A: The Alexander matrix associated to a Seifert matrix is $A - t A^T$.
The signature of the knot is the signature of the quadratic form associated to the matrix $A + A^T$, which is the case $t = -1$ in the previous matrix.
So: yes.
By the way, I highly recommend Livingston and Naik's Introduction to Knot Concordance for these topics. It's still unpublished, but I don't know of a better reference for all these knot invariants.
