The following problem has come up in my research and I don't have the tools (i.e., I don't know Algebraic Geometry, especially over $\mathbb{R}$) to solve it.
Consider two subsets $X$ and $Y$ of $M_n(\mathbb{R})^k$ with $k\geq 3$. The subset $X$ consists of all $k$ tuples of n x n matrices $(A_1,..., A_k)$ such that for any two $A_i$ and $A_j$, there is a vector $v_{ij}$ which is simultaneously an eigenvector for $A_i$ and $A_j$ (but perhaps with different eigenvalues).
The subset $Y$ consists of all those elements of $X$ such that $v_{ij}$ can be chosen independently from $i$ and $j$ - that is, if $(A_1,.., A_k)\in Y$ then all $k$ matrices have a common eigenvector. (Of course, when $k=1$ or $k=2$, $X = Y$, hence the above restriction on $k$).
Now, I have not been able to prove that $X$ is Zariski closed (though I have not tried that hard - it's not so important for my purposes), but I can prove that it's contained in a proper Zariski closed subset of $M_n(\mathbb{R})^k$ (thought of as $\mathbb{R}^{kn^2}$): we have $f_{12} = det(A_1A_2 - A_2A_1) = 0$ since $A_1A_2 v_{12} = A_2A_1 v_{12}$. Or, since we're worker over $\mathbb{R}$, we can put all the $f_{ij}$ into one big polynomial equation $\sum_{1\leq i < j\leq k} f_{ij}^2 = 0.$
What I'd like to know is
Is there a Zariski closed subset $F$ with $Y\subseteq F\subsetneq X$?
Said another way
Is there a polynomial which is simultaneously satisfied by all k-tuples of matrices sharing a common eigenvector but for which there are elements in $X$ which do not solve it?
Finally, in case it helps, the case I'm most interested in is $n=3$ and $k = 5$, but I imagine the choice of $n$ won't affect the answer greatly and $k=3$ probably contains all the insight necessary to tackle the larger $k$ values.
Thank you in advance for your help.