Let \begin{equation} P_{ij}^{k}(X_1,X_2,...,X_{2n-1},X_{2n})=\left\{ \begin{array}{ll} X_{2j-1}, & \hbox{ if $k=2i-1$;} \\ X_{2j}, & \hbox{ if $k=2i$;}\\ X_{2i-1}, & \hbox{ if $k=2j-1$;} \\ X_{2i}, & \hbox{ if $k=2j$;}\\ 0. & \hbox{ otherwise.} \end{array} \right. \end{equation}I wonder if the following is true: If $\{Q_{i,j}\}_{1\leq i\leq j\leq n}$ are polynomials in $X_1$, $X_2$,..., $X_{2n-1}$, $X_{2n}$ such that $$\tag{1} \sum_{1\leq i\leq j\leq n}Q_{i,j}P_{ij}^k=0\mbox{ for all }1\leq k\leq 2n, $$ then we must $Q_{i,j}=0$ for all $1\leq i\leq j\leq n$. I need this property to solve a system of linear first order PDEs, which is posted here.
If we let $Q_{i,j}=Q_{j,i}$, then $(1)$ can be written as $$\tag{2}Q_{1,1}X_1+Q_{1,2}X_3+\cdots+Q_{1,n}X_{2n-1}=0,\\ Q_{1,1}X_2+Q_{1,2}X_4+\cdots+Q_{1,n}X_{2n}=0,\\ Q_{2,1}X_1+Q_{2,2}X_3+\cdots+Q_{2,n}X_{2n-1}=0,\\ Q_{2,1}X_2+Q_{2,2}X_4+\cdots+Q_{2,n}X_{2n}=0,\\ \vdots\\ Q_{n,1}X_1+Q_{n,2}X_3+\cdots+Q_{n,n}X_{2n-1}=0,\\ Q_{n,1}X_2+Q_{n,2}X_4+\cdots+Q_{n,n}X_{2n}=0.$$
Here are some of my (fail) attempts: First I try to look at $$\tag{3} Q_{1}X_1+Q_{2}X_3+\cdots+Q_{n}X_{2n-1}=0,\\ Q_{1}X_2+Q_{2}X_4+\cdots+Q_{n}X_{2n}=0$$ and conclude that $Q_i$ must be zero, which is wrong by this answer.
Next I try to prove it by induction on $n$. Setting $X_{2n-1}=X_{2n}=0$ in $(2)$, we can use the induction assumption to conclude that $$Q_{i,j}(X_1,X_2,...,X_{2n-3},X_{2n-2},0,0)=0\mbox{ for all }1\leq i\leq j\leq n-1.$$ Similarly, we can set $X_{2k-1}=X_{2k}=0$ in $(2)$ and by the induction assumption we obtain \begin{equation} Q_{i,j}(X_1,...X_{2k-2},0,0,X_{2k+1},...,X_{2n})=0\mbox{ for }i\neq k\mbox{ or }j\neq k, \end{equation} where $k=1,2,..., n$. Therefore, I try to look at $(3)$ with the extra condition that $$Q_i(x_1,...,\underbrace{0}_{(2j-1)-\mbox{th}},\underbrace{0}_{2j-\mbox{th}},...,x_{2n})=0\mbox{ for all }j\neq i.$$ and conclude that $Q_i$ must be zero, which is still wrong by this answer.