The Grassmannian $G(n,k)$ is the space of $k$ dimensional linear subspaces of $\mathbb R^n$. The Plücker map $\pi: G(n,k) \to \mathbb R P^m, m = \binom nk -1,$ is defined by sending a linear space $L$ to the span of the vector of determinants of all $k\times k$ minors of a matrix that consists of basis vectors for $L$ in rows (since the minors of different bases are proportional this is well defined). I will refer to these minors as the basis's minors from now on.
I want to show that the image $\text{Im} \pi$ is a (smooth) $k(n-k)$-dimensional submanifold of $\mathbb R P^m$. To show that, I need to show that for all $\pi(L)\in \text{Im} \pi$ there is a chart $(U, \varphi)$ of $\mathbb RP^m$ such that $\pi(L) \in U$ and $U\cap \text{Im} \pi = \phi^{-1} (\mathbb R^{k(n-k)})$. In particular only $k(n-k)$ elements of $\phi (\pi (L))$ can be nonzero.
The charts I know for $\mathbb RP^m$ are $U_i = \{\text{span} (x_0, \dots, x_m): x_i \ne 0\}$ with $\phi_i: U_i \to \mathbb R^m, \text{span}(x_0, \dots, x_m) \mapsto (x_0, \dots, x_{i-1}, x_{i+1}, \dots, x_m)/x_i$. So, $\phi_i (\pi (L))$ consists of $\binom nk -1$ minors of a basis for $L$, each divided by the $\binom nk$'s minor.
Therefore, for $\text{Im} \pi$ to be a submanifold, every $k$-dimensional subspace $L$ of $\mathbb R^n$ should have at most $k(n-k) + 1$ nonzero minors, since otherwise there would be $k(n-k) + 1$ nonzero elements in $\phi_i (\pi (L))$.
However, the last claim is not true (I will post a counterexample if asked), so $\text{Im} \pi$ is not a submanifold of the projective space. This is false. Where is the flaw in my reasoning?