Let $M, N \subset \mathbb{R}^{n}$ be sets of matrices $3 \times 3$ with rank $1$ and $2$, respectively. Show that $M$ and $N$ are manifolds such that dim $M=5$ and dim $N=8$.
I have thought about using the determinant function $ \det: \mathbb{R}^{n^{2}} \to \mathbb{R}$ but i have problems when i deal with $f^{-1}(0)$.
On
For $A$ a $3\times 3$ matrix, we denote $A = \left(A_1, A_2, A_3\right)$. Let $M$ be matrices of $\mathcal{M}_{3\times 3}(\mathbb{R})$ of rank 1.
We have $M = M_1 \cup M_2 \cup M_3$, where $$M_1 = \left\{ A = (A_1, A_2, A_3) \mid A_1 \ne 0_{\mathbb{R}^3}, A_2 = \lambda_{2} A_1, A_3 = \lambda_{3} A_1 \text{ for } \lambda_2,\lambda_3\in\mathbb{R}\right\} $$
$$M_2 = \left\{ A = (A_1, A_2, A_3) \mid A_2 \ne 0_{\mathbb{R}^3}, A_1 = \lambda_{1} A_2, A_3 = \lambda_{3} A_2 \text{ for } \lambda_1,\lambda_3\in\mathbb{R}\right\} $$
$$M_3 = \left\{ A = (A_1, A_2, A_3) \mid A_3 \ne 0_{\mathbb{R}^3}, A_1 = \lambda_{1} A_3, A_2 = \lambda_{2} A_3 \text{ for } \lambda_1,\lambda_2\in\mathbb{R}\right\} $$
Every $M_i,\,i=1,2,3$ is a manifold of dimension $5$ ($M_i \cong \left(\mathbb{R}^3\setminus \{0_{\mathbb{R}^3}\}\right) \times \mathbb{R}^2$ and also $M_i\cap M_j$ is a manifold of dimension $5$. so $M$ is manifold of dimension $5$.
Let $\det: \mathbb{R}^9 \to \mathbb{R}$ be the determinant. we have $N = \det^{-1}(0) \setminus M$ and $\det^{-1}(0)$ is manifold of dimension $8$.
Let's refine the precedent answer.
Let $$M_{ij} = \left\{ A = (A_1, A_2, A_3)^T \in M\mid a_{ij} \ne 0\right\}$$ and $\varphi_{ij}: M_{ij} \to X_j\times \mathbb{R}^2$ be the chart, where $X_1 = \mathbb{R}^{\ast}\times\mathbb{R}^2$, $X_2 = \mathbb{R}\times\mathbb{R}^{\ast}\times\mathbb{R}$ and $X_3 = \mathbb{R}\times\mathbb{R}\times\mathbb{R}^{\ast}$. And $\varphi_{ij}(A) = \left(a_{i1}, a_{i2}, a_{i3}, (\frac{a_{kj}}{a_{ij}})_{k\ne i}\right)$.
N.B. why $\left(\frac{a_{kj}}{a_{ij}}\right)_{k\ne i}$: for exampel if $A = (A_1 A_2 A_3)^T\in M_{11}$ we know that $A_2 = \lambda_2 A_1,\, A_3 = \lambda_3 A_1$ for some $\lambda_1,\lambda_2\in\mathbb{R}$. Because $a_{11} \ne 0$, we deduce $\lambda_1 = \frac{a_{21}}{a_{11}}$; idem for $\lambda_2$.
It is clear that $X_j\times\mathbb{R}^2$ is an open of $\mathbb{R}^5$ and $\varphi_{ij}\circ\varphi_{lk}^{-1}: X_j\times\mathbb{R}^2\cap X_k\times\mathbb{R}^2 \to X_j\times\mathbb{R}^2\cap X_k\times\mathbb{R}^2$ is a $\mathcal{C}^{\infty}$ gluing map. so M is a manifold of dimension 5.
The same for N: Just consider $$N_{ij} = \left\{ A \in N\mid \text{the submatrix of } A \text{ resulting of removing ith row and jth column is of determinant non null} \right\}$$
For example $N_{33} = \left\{A\in N\mid a_{11}a_{22} - a_{21}a_{12}\ne 0\right\}$. Because $A= \left(A_1 A_2 A_3\right)^T\in N$, we know there exist $\lambda_{1},\lambda_{2}\in\mathbb{R}$, s.t. $A_3 = \lambda_1 A_1 + \lambda_2 A_2$. So in particular $\lambda_1,\lambda_2$ are solutions to the equation system: $$ a_{11}\lambda_1 + a_{12}\lambda_2 = a_{13}\\ a_{21}\lambda_1 + a_{22}\lambda_2 = a_{23} $$
so by Cramer rules we have: $$\lambda_1 = \frac{\det\begin{pmatrix}a_{13} & a_{12}\\a_{23} &a_{22}\end{pmatrix}}{\det\begin{pmatrix}a_{11} & a_{12}\\a_{21} &a_{22}\end{pmatrix}} = \frac{a_{13}a_{22} - a_{23}a_{12}}{a_{11}a_{22} - a_{21}a_{12}}$$
$$\lambda_2 = \frac{\det\begin{pmatrix}a_{11} & a_{13}\\a_{21} &a_{23}\end{pmatrix}}{\det\begin{pmatrix}a_{11} & a_{12}\\a_{21} &a_{22}\end{pmatrix}} = \frac{a_{11}a_{23} - a_{21}a_{13}}{a_{11}a_{22} - a_{21}a_{12}}$$
We define, for example, for $A =\left(A_1 A_2 A_3\right)^T\in N_{33}$ $$\varphi_{33}: N_{33} \to U_{33} \times \mathbb{R}^{2}\\ \varphi_{33}(A) = \left(A_1, A_3, \lambda_1, \lambda_2\right) = \left(A_1, A_2, \frac{a_{13}a_{22} - a_{23}a_{12}}{a_{11}a_{22} - a_{21}a_{12}},\frac{a_{11}a_{23} - a_{21}a_{13}}{a_{11}a_{22} - a_{21}a_{12}}\right) $$ where $U_{33} = \left\{(A_1 A_2)^T \in\mathcal{M}_{3\times 2}\mid a_{11}a_{22} - a_{21}a_{12} \ne 0\right\}$.
The same goes for all $N_{ij}$. It is clear that $U_{ij}\times\mathbb{R}^2$ is an open of $\mathbb{R}^8$ and the gluing maps: $$\varphi_{ij}\circ\varphi_{kl}^{-1}: U_{ij}\times\mathbb{R}^2\cap U_{kl}\times\mathbb{R}^2 \to U_{ij}\times\mathbb{R}^2\cap U_{kl}\times\mathbb{R}^2$$ are $\mathcal{C}^{\infty}$ maps.
This proves that $N$ is manifold of dimension $8$.