I am trying to show that the space of $2\times 2$ matrix with rank equals $1$ is a submanifold of $\mathbb{R}^4 - \{0\}$ whoose the dimension equals $3$. To do this, I have defined $\det : \mathbb{R}^4 \to \mathbb{R}$, then the result will follows from the fact that this submanifold is precisely $\det ^{-1}(0)$. But for this, I have to show that on this set, the derivative of $\det$ has constant rank.
How can I do this?
Hint: Consider the matrix: $$ M=\begin{bmatrix} a&b\\c&d \end{bmatrix}. $$
Then the $\det$ map is $(a,b,c,d)\mapsto (ad-bc)$. Therefore, $\det_\ast=\begin{bmatrix} d,-c,-b,a\end{bmatrix}$, which is a surjective map from $T_MM_{2,2}\rightarrow T_{\det(M)}\mathbb{R}$, except when $a=b=c=d=0$.
Therefore, in each neighborhood of a matrix $M$ with rank $1$, the submersion theorem applies. Since an open subset of a manifold is a manifold, we know that $M_{2,2}\setminus\{0\}$ is a manifold, and we can work on that (so that the zero matrix is not a problem).