How to understand why the set of nonsigular matrices is a differentiable submanifold of the the of matrices (of size $n$).
I thought of introducing, given a nonsingular matrix $A$, the mapping $f:X\mapsto \det(X)-\det(A)$. But from this on I do not know how to proceed.
Also, how to determine the tangent space at a point $A$?
On
The idea is that the singular matrices in $M_{m \times q}(\Bbb R)$ form a closed subset, as they constitute precisely the subset $\text{det}^{-1}(\{0\})$, and $\text{det}$ is a continuous function on $M_{m \times q}(\Bbb R)$. So the subset of nonsingular matrices is the open set $M_{m \times q}(\Bbb R) \setminus \text{det}^{-1}(\{0\})$, and any open subset of a Euclidean space is a submanifold of the same dimension (in this case, $mq$). For any nonsingular matrix $\mathbf{A}$, the tangent space to $\mathbf{A}$ in this open set is the same as $T_{\mathbf{A}} M_{m \times q}(\Bbb R)$, which can be identified with $\Bbb R^{mq}$.
More generally, let $\Sigma_k \subset M_{m \times q}(\Bbb R)$ be the space of matrices of rank $k$. For any matrix $\mathbf{A}$ in $\Sigma_k$, there is a $(k \times k)$-minor, let's call it $A$, in $\mathbf{A}$ which is nonsingular, and let the rest of the elements be arranged in three matrices $B_{k \times (q-k)}$, $C_{(m-k) \times k}$ and $D_{(m-k)\times(q-k)}$, where the arrangement is done by performing $k$ row operations $E_1$ and $k$ column operations $E_2$ so that in $E_2E_1 \mathbf{A}$, the minor $A$ is positioned at the upper left corner of the matrix, and $B, C, D$ are the other blocks. Since nonsingular matrices form an open subset, there is a neighborhood $U$ of $\mathbf{A}$ in $M_{m \times q}(\Bbb R)$ where for any $\mathbf{A}_0 \in U$, the $(k \times k)$-minor situated in the same position that $A$ is situated in $\mathbf{A}$ is also nonsingular. Therefore, the blocks $A, B, C, D$ of $\mathbf{A}$ are globally defined for every matrix in the open neighborhood $U$ of $\mathbf{A}$.
Let $\mathbf{X} \in U$ be arbitrary and let $A, B, C, D$ be the blocks of $\mathbf{X}$ as defined above. Notice that
$$\mathbf{X} \cdot \begin{pmatrix} I & -A^{-1}B \\ \mathsf{O} & I \end{pmatrix} = \begin{pmatrix} A & \mathsf{O} \\ C & D - CA^{-1}B \end{pmatrix}$$
Since the matrix we multiplied with is invertible, rank has remained unchanged. If $D - CA^{-1}B$ was nonzero we would have found a row in the bottom $(m-k)$-half of the matrix with a nonzero entry at one of the last $(q-k)$-th positions. This would then be linearly independent from the first $k$ rows of the matrix (as their last $(q-k)$ entries are all zero), which are themselves linearly independent by hypothesis on the minor $A$. That makes $\text{rank}(\mathbf{X}) \geq k+1$, absurd. Therefore $D - CA^{-1}B = \mathsf{O}$ is a necessary condition for $\text{rank}(\mathbf{X}) = k$. Once this condition holds the matrix above will have $k$ columns of vectors in $\Bbb R^m$ whose projection to the first $q$ coordinates are linearly independent, hence are themselves linearly independent.
So consider the map $F : U \to M_{(m-k)(q-k)}(\Bbb R)$ given by $F(\mathbf{X})= D - CA^{-1}B$. It's an easy check that $F$ is a smooth submersion, so $F^{-1}(\mathsf{O}) = U \cap \Sigma_k$ is a submanifold of codimension $(m-k)(q-k)$ in $U$, which proves that $\Sigma_k$ is a submanifold of $M_{m \times q}(\Bbb R)$.
This is not a closed, but a locally closed submanifold, of $M_{m \times q}(\Bbb R)$, and the closure $\overline{\Sigma_k}$ is precisely the subset of rank $\leq k$ matrices in $M_{m \times q}(\Bbb R)$. This can be seen by taking any matrix $\mathbf{M}$ with $\text{rank}(\mathbf{M}) = l < k$, which implies the $q$ column vectors $v_1, \cdots, v_q$ span a subspace of dimension $l$ in $\Bbb R^q$. Without loss of generality suppose $v_1, \cdots, v_l$ is a basis of this subspace. Then $\text{Span}(v_{l+1}), \cdots, \text{Span} (v_{l+k})$ belong to this $l$-dimensional subspace; by a small perturbation in $\Bbb R^q$ we can make them transverse to this $l$-dimensional subspace as well as with themselves. Let $w_{l+1}, \cdots, w_{l+k}$ be the modified vectors; the matrix formed by the column vectors $v_1, \cdots, v_l, w_{l+1}, \cdots, w_{l+k}, v_{l+k+1}, \cdots, v_q$ is a perturbation of $\mathbf{M}$ which has rank $k$ (Note: if $k > q$, do the same argument with the $m$ rows). However, rank can never go down, as transverse subspaces remain transverse under small perturbations.
Therefore, $\Sigma_k \subset M_{m \times q}(\Bbb R)$ are locally closed submanifolds such that $\overline{\Sigma_k} = \Sigma_0 \cup \cdots \cup \Sigma_{k-1}$. This kind of iterative structure is called a stratification; indeed, $\Sigma_k$'s constitute stratums for a Whitney stratification of $M_{m \times q}(\Bbb R)$. But this was a long digression.
It is an open subspace of a vector space, $Gl(n,\mathbb{R})=det^{-1}(\mathbb{R}-\{0\}$. Here the vector space is the space of $n\times n$-matrices so it is a submaniflod with one chart which is the embedding map $Gl(n,\mathbb{R})\rightarrow M(n,\mathbb{R})$.