Everybody knows that a square matrix $A$ has the same eigenvalues as
$A^T$. And it is clear that if $A^T=BAB^{-1}$ then $B$ maps eigenvectors
of $A$ to those of $A^T$. But I have not found any discussion of the
benefits of knowing $B$. Perhaps it is unusual to know $B$ exactly,
without having analyzed $A$ completely. But I have some examples where
this is known. 


It seems at least to be nontrivial information. For example, with $A$ real
and $B$ real and
symmetric, you know an indefinite inner product
$\langle x,y\rangle = x^HBy$
with respect to which $A$ is symmetric. This gives certain orthogonality
relations (for example non-real eigenvalues have null eigenvectors) but it doesn't
seem to lead to anything quantitative about eigenvalues.


The worst case may be for symmetric $A$, $A^T=IAI^{-1}$ tells you nothing.


So the question is



> 
> Have examples been studied, where knowing $B$ in $A^T=BAB^{-1}$ helped to find eigenvalues of $A$? 
> 
> 
> 



Does knowing a conjugation of A to A^T determine eigenvalues of A?
Generally, over the algebraically closed field, say over $\mathbb{C}$, $A^T$ and $A$ are always conjugate, because they have the same Jordan form. So such $B$ always exists. But the knowledge of the concrete $B$ satisfying to this relation may clarify what the eigenspaces for $A$ are, because ${B^T}B$ always commutes with your A. As previously was mentioned, this condition $A^T=BAB^{-1}$ doesn't determine the eigenvalues of A uniquely, so it may help only with eigenspaces. And the amount of information it gives you depends on this particular $B$. For $B=I$ it gives you nothing, as you mentioned above.


