Show the Grassmannian is a smooth manifold (using dummy definition of smooth manifold) We received the following problem in my Differential Geometry class:
Suppose $0\leq k \leq n$ are integers. Let $G(k,n)$ be the collection of orthogonal projections $T: \mathbb{R}^n \to \mathbb{R}^n$ with rank $k$. Identifying the collection $L(\mathbb{R}^n , \mathbb{R}^n)$ of linear transformations $\mathbb{R}^n \to \mathbb{R}^n$ with the collection $M_{n \times n}$ of $n \times n$ matrices, show that $G(k,n) \subset M_{n\times n} \simeq \mathbb{R}^2$ is a smooth manifold.
Here our (provisional) definition of a $k$-manifold is
$M \subset \mathbb{R}^n$ is a $k$-manifold if for each point $p\in M$ there exists a neighborhood $U$ of $p$ and $I = \{i_1, \dotsc, i_k\} \subset \{1,\dotsc,n\}$ such that $U \cap M$ is the graph of a $C^\infty$ function $f: V \to \mathbb{R}^{I^c}$, where $V \subset \mathbb{R}^I$.
Now I know the Grassmannian should have dimension $k(n-k)$ (see this question). That means that out of the conditions $F: \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$ requiring $\rho$ to be an orthogonal projection of rank $k$, I need to find a submatrix of $DF$ of dimension $n^2 - k(n-k) = n^2 - kn +k^2$ which is non-singular. 
The requirements are 1. $\rho^2 = \rho \; $ 2. $\rho^T = \rho \;$ 3.  $\operatorname{rk}{(\rho)}=k$. 
So
$$a_{ij} = a_{ji} \tag{1}$$
$$\sum_{k=1}^n a_{ik}a_{kj} = a_{ij} \tag{2}.$$
I need to find a clever way to express the last one as an equation, and to show that at each point there is the necessary submatrix. Any ideas?
 A: Here's an easier way that I found to do it, after a year. Unfortunately it doesn't use the "dummy" definition of manifold, as originally intended.
Define an index set $\Omega:= \{\omega \in \mathbb{Z}^n \mid 1 \leq \omega_1 < \omega_2 < \dotsb < \omega_k \leq n \}$.
 Start with the open set of $n \times k$ matrices of rank $k$ (open by the equivalent condition $\det A^TA \neq 0$, since $\operatorname{Ker}(A^TA) = \operatorname{Ker}(A)$). Now quotient by the relation $A \sim B$ if $\operatorname{Rg}(A)=\operatorname{Rg}(B)$. Each matrix $A$ has an invertible $k \times k$ submatrix; that is, $C_{\omega}A \in \text{GL}_k(\mathbb{R})$ for some $\omega \in \Omega$, where $C_\omega$ chooses the rows of $A$ corresponding to index $\omega$. Now cover the quotient space by the images of $V_\omega$, where $V_{\omega}$ is the open subset of matrices of rank $k$ whose $\omega$ rows form a non-singular matrix. On $\overline{V_{\omega}}$ parameterize by the $k(n-k)$ entries left over when we turn the $\omega$ rows into the identity matrix, that is, map 
 $$ A \mapsto C_{\omega^{c}}A(C_\omega A)^{-1}. \tag{1} $$ 
 This is well-defined on equivalence classes, since if $B= AU$, then $AU(C_\omega AU)^{-1} = AUU^{-1}(C_{\omega}A)^{-1} = A (C_{\omega}A)^{-1}$.
The map (1) $= \phi \circ \pi $ is clearly continuous. But $\phi \circ \pi$ is continuous iff $\phi$ is continuous by property of quotients. The inverse map is the map $C_{\omega^c}A \mapsto A \mapsto \bar{A}$. The first map is clearly continuous, and we know the second map is continuous.  
If $\overline{A} \in \overline{V_{\omega}}\cap \overline{V_{\omega'}}$, where $C_{\omega}A = \text{Id}$, then a transition map is of the form $C_{\omega^c}A \mapsto A \mapsto C_{\omega'^c}A(C_{\omega'}A)^{-1} $, which is clearly $C^{\infty}$. 
Note: This method generalizes the standard way to show that projective space is a smooth manifold.
