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:
> $$\sum\_{1\leq i<j\leq n}a\_{ik}a\_{jk}x\_{ij}=0\ \text{for all invertible matrix}\ A=(a\_{ij})$$
> 
> 
> 



> 
> 1. Does this imply that $x\_{ij}=0$ for all $1\leq i<j\leq n$?
> 2. Is it possible to write the equality above using some quadratic form?
> 
> 
> 



Degenerate representation
Suppose $V = \mathbb{R}^n$ has a basis $(e\_1,\dots,e\_n)$. Your assumption is that you have a family of linear maps $\lambda\_1,\dots,\lambda\_m \in V^\*$ which are defined such that $\lambda\_r(e\_i) = \lambda\_{ir}$ for any $1 \leqslant r \leqslant m$. Working inside the algebra of all functions $\mathrm{Fun}(V,\mathbb{R})$, with multiplication given by pointwise evaluation, we have a function


$$q\_{rs} = \lambda\_r\lambda\_r - \lambda\_s\lambda\_s$$


for any $1 \leqslant r,s \leqslant m$. This is, indeed, a quadratic form on $V$.


If $(e\_1^\*,\dots,e\_n^\*)$ is the dual basis of $V^\*$ then $\lambda\_r = \sum\_{i=1}^n \lambda\_{ir}e\_i^\*$ and


$$\lambda\_r\lambda\_r = \sum\_{i=1}^n \lambda\_{ir}^2e\_i^\* + 2\sum\_{1 \leqslant i < j \leqslant n} \lambda\_{ir}\lambda\_{jr}e\_i^\*e\_j^\*$$


Your assumption that $\lambda\_{ir}^2 = \lambda\_{is}^2$ for any $1 \leqslant r,s \leqslant m$ means that


$$q\_{rs} = 2\sum\_{1 \leqslant i < j \leqslant n} (\lambda\_{ir}\lambda\_{jr}-\lambda\_{is}\lambda\_{js})e\_i^\*e\_j^\*$$


As you do, we could divide by $2$ to get a quadratic form $q\_{rs}'$ so that $q\_{rs} = 2q\_{rs}'$.


If $g \in \mathrm{End}(V)$ then $ge\_k = \sum\_{i=1} a\_{ki}e\_i$ for some $a\_{ki} \in \mathbb{R}$ and we have $e\_i^\*(ge\_k)e\_j^\*(ge\_k) = a\_{ki}a\_{kj}$ so you have


$$q\_{rs}'(ge\_k) = \sum\_{1 \leqslant i < j \leqslant n} a\_{ki}a\_{kj}(\lambda\_{ir}\lambda\_{jr}-\lambda\_{is}\lambda\_{js})$$


So your condition reads that $q\_{rs}'(ge\_k) = 0$ for all $g \in \mathrm{GL}(V)$.


Now let $q : V \to \mathbb{R}$ be a quadratic form. Your question amounts to asking whether $q = 0$ given that $q(ge\_i) = 0$ for all $g \in \mathrm{GL}(V)$ and $1 \leqslant i \leqslant n$. If we were working over $\mathbb{C}$ then we could show this by viewing $V$ as an affine space and using that $q$ is a polynomial function. So as $\mathrm{GL}(V)$ is dense in $\mathrm{End}(V)$ we must have $q = 0$. Probably there is an equivalent argument over $\mathbb{R}$ but I'll leave this to someone else.


One can instead argue as follows. For any $i \neq j$ we see that the map $\tau\_{ij} \in \mathrm{End}(V)$ given by $\tau\_{ij}e\_k = e\_k + \delta\_{ki}e\_j$ is invertible and $\tau\_{ij}e\_i = e\_i+e\_j$. Hence $q(e\_i+e\_j) = 0$ for any $1 \leqslant i,j \leqslant n$ (if $i=j$ then this is $q(2e\_i) = 4q(e\_i) = 0$).


Now let $\beta(u,v) = q(u+v) - q(u) - q(v)$ be the symmetric bilinear form determined by $q$. Then for any $1 \leqslant i,j \leqslant n$ we have


$$\beta(e\_i,e\_j) = q(e\_i+e\_j)-q(e\_i) - q(e\_j) = 0$$


Using the bilinearity we see that $\beta(u,v) = 0$ for any $u,v \in V$ which shows that $q(v) = \frac{1}{2}\beta(v,v) = 0$ for any $v \in V$. Thus $q = 0$ is identically zero.


