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. There is an orthonormal basis $d_1,\ldots,d_{2m}$ of $\mathbb{R}^{2m}$ such that for all $i=1,\ldots,n$ and $j=1,\ldots,m$
$$\rho(e_i)(d_{2j-1})=\lambda_{ij}d_{2j}\ \text{and}\ \rho(e_i)(d_{2j})=-\lambda_{ij}d_{2j-1}$$
for some real numbers $\lambda_{ij}$.
The **degeneracy** property of this representation is as follows
\begin{equation}\label{deg}\tag{1}
\lambda_{ki}^2=\lambda_{kj}^2\ \text{for all}\ k=1,\ldots,n\ \text{and for all basis of}\ \mathbb{R}^n.
\end{equation}
If $v_k=\sum_{p=1}^na_{pk}e_p$ $k=1,\ldots,n$ is another basis of $\mathbb{R}^n$, then
\begin{align*}
\text{$\eqref{deg}$}& \Longleftrightarrow \left(\sum_{p=1}^na_{pk}\lambda_{pi}\right)^2-\left(\sum_{p=1}^na_{pk}\lambda_{pj}\right)^2=0\ \text{for all}\ k=1,\ldots,n\\
& \Longleftrightarrow \sum_{p=1}^na_{pk}^2(\lambda_{ki}^2-\lambda_{kj}^2)+2\sum_{1\leq p<q\leq n}a_{pk}a_{qk}\left(\lambda_{pi}\lambda_{qi}-\lambda_{pj}\lambda_{qj}\right)=0\ \text{for all}\ k=1,\ldots,n\\
& \Longleftrightarrow \sum_{1\leq p<q\leq n}a_{pk}a_{qk}\left(\lambda_{pi}\lambda_{qi}-\lambda_{pj}\lambda_{qj}\right)=0\ \text{for all}\ k=1,\ldots,n
\end{align*}

So I am studying the following question: $$\sum_{1\leq i<j\leq n}a_{ik}a_{jk}x_{ij}=0\ \text{for all invertible matrix}\ A=(a_{ij})$$

- Does this imply that $x_{ij}=0$ for all $1\leq i<j\leq n$?
- Is it possible to write the equality above using some quadratic form?