I study real $4\times 4$ matrices, which have the following form:
$$ A\left(\lambda\right) = \begin{pmatrix} \epsilon_{1}\left(\lambda\right) & B & \Delta & 0 \\ B & \epsilon_{2}\left(\lambda\right) & 0 & \Delta \\ \Delta & 0 & \epsilon_{3}\left(\lambda\right) & B \\ 0 & \Delta & B & \epsilon_{4}\left(\lambda\right) \end{pmatrix}\text{.} $$
with $B\in\mathbb{R}$, $\Delta\in\mathbb{R}$, $\epsilon_{1}\left(\lambda\right) = \epsilon_{4}\left(-\lambda\right)\in\mathbb{R}$ and $\epsilon_{2}\left(\lambda\right) = \epsilon_{3}\left(-\lambda\right)\in\mathbb{R}$. Furthermore, the matrix $A$ is hermitian and for the eigenvalues it is valid that $E\left(\lambda\right) = E\left(-\lambda\right)$.
I am interested in finding the transformation $U\left(\lambda\right)$, which transforms $A\left(\lambda\right)$ in the diagonal form. I know that $A\left(\lambda\right) = A^{T}\left(\lambda\right)\in\mathbb{R}^{4\times 4}$, thus $U\left(\lambda\right)\in O\left(4\right)$.
My question is: can I say something about the transformation (e.g. that $U\left(-\lambda\right) = U^{T}\left(\lambda\right)$ or something else) without calculating explicitly the eigenvectors of $A\left(\lambda\right)$ and finding with this the $U\left(\lambda\right)$?