Is $ G_{n,k} = \{ M \in M_{n\times n}(\mathbb{R}) ; \hspace{.1cm} \text{rank} (M) = k , M$ is symmetric and $ M ^2= M \}$ a manifold? I'm having trouble to prove that the intersection of the symmetric matrices , the set of projections (matrices who $M^2 = M$) and the matrices with constant rank $k$ is a manifold. Can anyone help me? Unfortunately I could not make much progress.
In other words I need to prove the following proposition: Let $M_{n\times n} (\mathbb{R})$ the set of the real matrices $n \times n$, considere $k \in \{1,2,3, ..., n \}$, then   $ G_{n,k} = \{ M \in M_{n\times n}(\mathbb{R}) ; \hspace{.1cm} \text{rank} (M) = k ,  M$   is symmetric and $ M ^2= M  \}$ is a smooth manifold?
 A: It suffices to note that the function $f:\mathrm{Sym}_{n \times n} \to \Bbb R^2$ given by
$$
f(X) = (\operatorname{tr}([X^2 - X]^T[X^2 - X]),\operatorname{tr}(X))
$$
is smooth (since the components are polynomials on matrix-entries) and regular.  So, the set $\{X : f(X) = (0,k)\}$ is a manifold.
A: We can identify the space $G_{n,k}$ with the Grassmannian $Gr(n,k)$ of  $k$-dimensional linear subspaces of $\mathbb{R}^n$. By the Theorem of constant rank, a smooth neighborhood retract is automatically a submanifold. We can easily construct such a smooth retraction $U\rightarrow G_{n,k}$ for some $U$ in $\mathbb{R}^{\binom{n}{2}}$. For details see here, between Lemma $2.3$ and $2.4$.
A: A symmetric projection matrix is the same as an orthogonal projection onto its image. (If $v \in im T ^{\perp}$ ,then $<w,Tv> = <Tw,v> = 0$ for all $w$, so $Tv = 0$. If $Tv = 0$ for some $v$, the same computation shows it is orthogonal to the image.) If you fix the rank to be $k$, you can identify your matrices with $k$ dimensional subspaces of $\mathbb{R}^n$. This is the Grassmannian of $k$ -planes, which is "well known" to be a manifold, though it requires some work.
