discriminant matrix of ideal in number field 
*

*Let $K$ be a number field. Take $b_1,\ldots,b_n \in \mathcal{O}_K$. 
What I call the discriminant matrix is $$M_b = \left(\begin{array}{cccc}
\sigma_1(b_1) & \sigma_1(b_2) &\cdots & \sigma_1(b_n) \\
\sigma_2(b_1) & \ddots & & \vdots \\
\vdots & & \ddots & \vdots \\
\sigma_n(b_1) & \cdots & \cdots & \sigma_n(b_n)
\end{array}\right) \qquad \in \mathbb{C}^{n \times n}$$
where $ n=[K:\mathbb{Q}]$ and $\sigma_1,\ldots,\sigma_n$ are the embeddings $K \to\mathbb{C} $ (the Galois group of the normal closure).

*Usually, we take $b$ to be an integral basis of $\mathcal{O}_K$. But I'm insterested in the case $b$ is a basis of an ideal $I \subset \mathcal{O}_K $ (seen as a free $\mathbb{Z}$-module).  
Question :   
How to interpret the matrix $M_b$ ? Do  we know its eigenvalues ? Why is determinant tell us something about $I$ and $\mathcal{O}_K$ ? As a linear operator $\mathbb{C}^n \to \mathbb{C}^n$, it sends the lattice $\mathbb{Z}^n $ to the lattice $\sigma(I)$. 
How to interpret $M_b^{-1}$ ? Can we compute it ? How is it related to the ideal $J \subset \mathcal{O}_K$ such that $IJ = (\delta)$ ? (it seems that $M_b^{-1} = \frac{1}{\delta} M_a$ where $a_1,\ldots,a_n$ is some basis of $J$)
If it is easier, take $n=2$ or $K$ monogenic, or anything making it easier to visualize. This is indeed the same question as thisone.
 A: I'm going to try to answer some of your questions from a beginner point of view with things you probably already know.
- How to interpret $M_b^{-1}$ ?  Can we compute it ?
As you pointed out  here and in the your previous post, this is related to the different ideal and the concepts that leads to its definition ( I think this is a good exposition of it ).
In particular, let $b^\vee=\{ b_1^\vee, \ldots,  b_n^\vee\}$ be the dual basis of $b$, i.e.,  the only $\mathbb{Q}$-basis of $K$ such that $\text{Tr}_{K/\mathbb{Q}}(b_i\, b_j ^\vee)=\delta_{ij}$ (kronecker delta). Then,
$I ^\vee:=\{x \in K \mid  \text{Tr}_{K/\mathbb{Q}}(x I) \subset\mathbb{Z} \} $ 
is a $\mathcal{O}_K$-submodule of $K$, and since for any $x\in K $ we have
$x=\displaystyle\sum_{i=1}^{n}\text{Tr}_{K/\mathbb{Q}}(x\,b_i) \cdot b_i ^\vee $
then $I ^\vee$ is exactly the $\mathbb{Z}$ span of $b^\vee$ in particular is finitely generated over $\mathcal{O}_K$ so is a fractional ideal. 
Now is easy to verify $M_b^{-1}=M_{b^\vee}$, since
$(\, \sigma_i(b_j) \,)\cdot (\, \sigma_i(b_j ^\vee) \,) = (\displaystyle\sum_{k=1}^{n} \sigma_k(b_i) \sigma_k(b_j ^\vee) )=(\displaystyle\sum_{k=1}^{n} \sigma_k(b_i b_j ^\vee) )=(\text{Tr}_{K/\mathbb{Q}}(b_i\, b_j ^\vee))=(\delta_{ij})$
Of course you can multiply by some $r \in \mathbb{N}$ so that $J= r I^\vee$ is an integral ideal and then $a=r \cdot b^\vee$ is a basis of $J$ such that
$M_b^{-1}=  \frac{1}{r} M_{a}$.
We also have the identity
$I \cdot I^\vee= \mathcal{O}_K ^\vee $
Where $\mathcal{D}_K:= (\mathcal{O}_K ^\vee)^{-1}$ is by definition the different ideal, which as mention here is not always principal, but in the case $\mathcal{O}_K:=\mathbb{Z}[\alpha]$ is monogenic $\mathcal{D}_K=f'(\alpha) \mathcal{O}_K$ where $f(x)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$.
So your equation would read $I \cdot J =(\delta)$ where $\delta=r/f'(\alpha)$, but in general is just $I \cdot J= r \, \mathcal{O}_K ^\vee$.
-How to interpret the matrix $M_b$ ? 


*

*Why its determinant tell us something about $I$ and $\mathcal{O}_K$?:


Well $M_b^{2}=T_b $ where $T_b=( \text{Tr}_{K/\mathbb{Q}}(b_i \, b_j))$ is the 
    matrix of traces, so $ \text{det}(M_b)^{2}=\text{det}(T_b)=\text{Disc}(I/\mathbb{Z})=[\mathcal{O}_K:I]^{2} \, \text{Disc}(\mathcal{O}_K)=N(I)^{2}\, \text{Disc}(\mathcal{O}_K)$. So I guess its square tell us what primes $p\in \mathbb{Z}$ either ramify or have common factors with $I$.


*

*Do we  know its eigenvalues ?:


Let $\{ \lambda_1, \ldots ,\lambda_n \}$ be the eigenvalues of $M_b$. Since $M_b^{2}=T_b$ then $\{ \lambda_1, \ldots ,\lambda_n \}$ is the 
            square of the set  $\{ \mu_1, \ldots ,\mu_n \}$ of eigenvalues of $T_b$.
In particular we have things like
$\lambda_1^{2}+ \ldots +\lambda_n^{2}=\text{Tr}_{K/\mathbb{Q}}
     (b_1^{2}+\ldots+b_n^{2} )  $
$\lambda_1^{2}\cdot \ldots \cdot \lambda_n^{2}=N(I)^{2}\, \text{Disc}(\mathcal{O}_K)$
And since the characteristical polynomial $P_b(x)$ of $T_b$ has integer coefficients and each $\lambda_i$ satisfies $P_b(\lambda_i^{2})=0$  we know that they are all algebraic integers of degree at most $2n$.
Other than that at least I don't see any other immediate properties  of  $\{ \lambda_1, \ldots ,\lambda_n \}$ or the extention $\mathbb{Q}(\lambda_1, \ldots ,\lambda_n )$. I hope you get more comments on that question or that you  work out something yourself.
