It seems easy but I can't prove it. Can anyone give proof or reference?
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$\begingroup$ Almost any textbook that covers Alexander modules should do. $\endgroup$– Ryan BudneyCommented Jan 18, 2012 at 21:12
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If $A$ is a Seifert matrix for $K$ and $\omega \in \mathbb{C}$ has norm 1, then the Tristram-Levine signature $\sigma_\omega(K)$ is the signature of the matrix
$(1-\omega)A + (1-\bar{\omega})A^T = (1-\bar{\omega})(A^T - \omega A),$
which jumps when some eigenvalue of $A^T - \omega A$ crosses zero (i.e. changes sign). At these values of $\omega$ the product of the eigenvalues, which is $\det(A^T-\omega A) = \Delta_K(\omega)$, must therefore be zero.
answered Jan 18, 2012 at 17:09