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 $\ldots$



Geometry of Hermitian rank $\leq r$ matrices
Let $n=4$, $r=3$. Let $E = \mathbb{R} \oplus M\_3^{sa}(\mathbb{C})$ (viewed as $4\times 4$ matrices which are $0$ in the first row and column except for the $(1,1)$ entry). Then the (real) codimension of $E$ in $M\_4^{sa}(\mathbb C)$ is $6<9=r^2$. Let $F\subset M\_4^{sa}(\mathbb{C})$ be the space of all matrices with $0$ in the $(1,1)$ entry. Clearly $V$ is not contained in $F$. 


Now consider the matrix $X= 0\oplus I\_3$. We have $X\in V\cap F$, and small perturbations $Y$ of $X$ within $V$ must of course also lie in $E$, and thus have the form $Y=c\oplus (I\_3+A)$ where $c\in \mathbb R$ and $A\in M\_3^{sa}(\mathbb C)$, and we have $\|Y-X\|=\text{max}(|c|, \|A\|)$. But now if $\|Y-X\|<1$, then $\|A\|<1$, so $I+A$ has rank 3 which forces $c=0$ and thus $Y\in F$. 


