For the vector space $M_{n,n}(\mathbb{C})$ of $n\times n$ matrices we know that the subset
$$M_{2r}:= \{A\in M_{n,n}(\mathbb{C}) \mid rk{A}=2r \}$$$$M_{2r}:= \{A\in M_{n,n}(\mathbb{C}) \mid \mbox{rank} (A) = 2r \}$$
is a manifold of dimension $2n(2r)-(2r)^2$ (see this question).
If I now take a look onat Hermitian $n\times n$$n \times n$ matrices of rank $2r$ with exactexactly $r$ positive and $r$ negative eigenvalues, how can I determine the dimension of this manifold?