Let $M_n^{sa}$ be the space of $n\times n$ complex Hermitian matrices, let $r < n$, and let $E$ and $F$ be (real) linear subspaces of $M_n$ with ${\rm codim}(E) < r^2$ and ${\rm codim}(F) = 1$. Let $V$ be the set of matrices in $E$ whose rank is at most $r$ and suppose $V$ is not contained in $F$. Is $V \cap (M_n\setminus F)$ dense in $V$?

For this to fail, there would have to be some open subset of $V$ which is contained in $F$, while not all of $V$ is contained in $F$.

I assume the answer is yes, but I think this is an algebraic geometry question and I know next to nothing about the subject $\ldots$