Classification of bilinear forms: operator $A^{-1} A^T$ for bilinear form $A$

269 Views Asked by At

I would like to understand a classification of non-degenerate (not necessary symmetric or skew-symmetric) bilinear forms over an algebraically closed field via an operator $\kappa=A^{-1} A^T$ for a bilinear form $A$, or invariantly, $A(x,y)=A(y,\kappa x)$.

I have heard on a seminar that there is only one series $U_i$ of unipotent bilinear forms, and some other series we are not interested in, and that it directly follows from Jordan normal form for the operator $\kappa$. Could you help me with proof or references?

For $BAB^T$, the second is $B^{-T}A^{-1}A^TB^T=B^{-T}\kappa B^T$, so it is indeed an operator. The operator $\kappa=E$ iff $A$ is symmetric, and $\kappa=-E$ iff $A$ is skew-symmetric. For a direct sum $A \oplus B$, $\kappa(A \oplus B)=\kappa(A) \oplus \kappa(B)$, so we need indecomposable matrices. In dimension $2$, $A$ can be written as $$\begin{bmatrix}1 & k \\ 0 & 1\end{bmatrix} \text{, then } \kappa=\begin{bmatrix}1 - k^2 & -k \\ k & 1\end{bmatrix},$$ so its characteristic polynomial $x^2+(k^2-2)x-1$ has two different roots. Alas, I am stuck.