Let $\rho : \mathbb{R}^n\to \mathfrak{so}(2m)$ be a faithful representation of the commutative Lie algebra $\mathbb{R}^n$ into the Lie algebra of skew-symmetric matrices (a Lie algebra morphism). Since $\rho(\mathbb{R}^n)$ is contained in the maximal abelian subalgebra of $\mathfrak{so}(2m)$, then $n\leq m$ and there is an orthonormal basis $d_1,\dotsc,d_{2m}$ of $\mathbb{R}^{2m}$ such that for all $i=1,\dotsc,n$ and $j=1,\dotsc,m$ $$\rho(e_i)(d_{2j-1})=\lambda_{ij}\,d_{2j}\ \text{and}\ \rho(e_i)(d_{2j})=-\lambda_{ij}\,d_{2j-1}$$ for some real numbers $\lambda_{ij}$. In addition, we assume that for all $i=1,\dotsc,m$, there is $k=1,\dotsc,n$ such that $\lambda_{ki}\neq0$.

The **degeneracy** property of this representation is as follows
\begin{equation}\label{deg}\tag{1}
\exists i<j\in\{1,\dotsc,m\}, \forall k\in\{1,\dotsc,n\}, \text{s.t for all basis of}\ \mathbb{R}^n,\ \lambda_{ki}^2=\lambda_{kj}^2.
\end{equation}
This is a well defined notion (coordinate free), according to Degenerate representation.

I would like to study those representations for which \begin{equation}\label{eigen}\tag{2} \exists \ell,i<j\in\{1,\dotsc,m\}, \forall k\in\{1,\dotsc,n\}, \text{s.t for all basis of}\ \mathbb{R}^n,\ \lambda_{k\ell}^2=(\lambda_{ki}\pm\lambda_{kj})^2, \end{equation} i.e. $\lambda_{k\ell}^2$ is an eigenvalue of the matrix: $$A=\begin{pmatrix} \left(\lambda_{ki}^2+\lambda_{kj}^2\right)&-2\lambda_{ki}\lambda_{kj}\\ -2\lambda_{ki}\lambda_{kj}&\left(\lambda_{ki}^2+\lambda_{kj}^2\right) \end{pmatrix}=\begin{pmatrix} -\lambda_{kj}&\phantom{-}\lambda_{ki}\\ \phantom{-}\lambda_{ki}&-\lambda_{kj} \end{pmatrix}^2.$$

Is this well defined (coordinate free)? In other words, is there a particular representation in this family for which $\eqref{eigen}$ is satisfied regardless of the chosen basis of $\mathbb{R}^n$?