Let $K$ be a number field of degree $n$, $\sigma_1 , \ldots , \sigma_n$ be the distinct embeddings of $K$ into $\mathbb C$, and define $\Delta(x_1, \ldots , x_n) = \det(\sigma_i (x_j)^2) = \det(\mathrm{Tr}_{K/\mathbb Q} (x_i x_j) )$.
Apparently, if x_i' = \sum_{j=1}^n a_{ij} x_j, where $a_{ij} \in \mathbb Q$ and $A = a_{ij}$, then \Delta(x_1', \ldots , x_n ') = (\det A)^2 \Delta(x_1, \ldots ,x_n)
I can't see why this is true (though I think it should be fairly easy). I'd appreciate if someone could explain.
Thanks