# Geometry of Hermitian rank $\leq r$ matrices

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 $\ldots$

• What if $F=E$ ? Dec 8, 2015 at 7:35
• Wouldn't then $V$ be contained in $F$?
– Dirk
Dec 8, 2015 at 7:44
• Right, I missed that bit... Dec 8, 2015 at 19:14
• @Nik You might want to add the tag "real algebraic geometry", since the subspace of Hermitian matrices is not a complex subspace (its defining equations involve conjugates). If everything was "complex" then I think $V$ would be irreducible and what you're asking would be true, but since the situation becomes "real", $V$ might be disconnected, and then I'm not sure. Dec 8, 2015 at 20:00
• @MattiaTalpo: did it. Yes, I was hoping I could just solve this by quoting some standard algebraic geometry, but since $V$ isn't a complex variety that seemed unlikely ... Dec 8, 2015 at 20:07

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 $Y\in F$.
• Yes, that is correct. But I have a condition on dimensions that I didn't think I would need which I would now like to add. Can your counterexample be adapted to give $E$ smaller codimension? Dec 8, 2015 at 21:27
• Yes, I think it's possible to get the codimension condition on $E$; I've edited the answer accordingly. Dec 8, 2015 at 21:47