Open cover of a grassmannian This is a "please check my understanding" type question.
I'm reading Andreas Gathmanns lecture notes on Algebraic geometry, found here, and I'm trying to get a better understanding of construction 8.15 on page 70.
Gathmann writes: Let $U_0 \subset G(k,V) \subset \mathbb{P}^{{n \choose k}-1}$ be an open affine subset where $e_1 \wedge ... \wedge e_k \neq 0$. By a remark elsewhere, the subspace spanned by $v_1...v_k$, denoted $L$ = Lin$\{v_1...v_k\}$ will be in $U_0$ iff the $k \times n$ matrix formed by the vectors $v_1...v_n$ is of the form $A = (B \mid C)$ with $B$ invertible $k \times k$. We can reduce $B$ to a unit matrix leaving a matrix $(E_k\mid D)$, this $D$ is unique and so we get a bijection
$$f:\{k\times (n-k)-\text{matrices}\} \to U_0 \\
D \mapsto \text{the space spanned by the rows of $(E_k\mid D)$} 
$$
The whole point here is to justify that $G(k,n)$ has dimension $k(n-k)$.
I'm wondering about the case when $U_0$ is not enough, so I need more affine open sets. Heres an example: 
Take, in $G(2,4)$, the space spanned by $v_1 = e_1 + 2e_2$ and $v_2 = 3e_1 + 4e_2$. Call this space $W$. Let's denote the affine open set where $e_1 \wedge e_2 \not= 0$ by $U_{12}$. Then $D$ in the map above would be the zero matrix.
Take the space $W'$ spanned by  $v_1 = e_1 + 2e_2 + e_3$ and $v_2 = e_1 + 2e_2 + 2e_3$. Now it seems that $W'$ is not in $U_{12}$, but in $U_{13}$ or $U_{23}$. 
It then seems to be that I would need all six affine sets $U_{i,j}$ to completely cover the the grassmannian, is this true? or can I get by with less? How many? Ultimately, does it matter?
And finally, did I understand everything properly? Is it true to say that these are the open sets that make the grassmannian into a manifold of dimension $k(n-k)$?
Any addition explanations and theory that you want to provide is most welcome!
 A: 
I'm wondering about the case when $U_0$ is not enough, so I need more affine open sets.

Contrary to the name, the construction in the notes just seems to be showing that $G(k, n)$ has an open subset $U_0$ isomorphic to $\mathbb{A}^{k(n-k)}$.  $U_0$ here is just the big Schubert cell, so $G(k, n) \setminus U_0$ is the union of all the other Schubert cells (or equivalently, of all the other Schubert varieties).
Since $G(k,n)$ is projective and $U_0$ is affine, the only time $U_0$ alone gives an open cover of $G(k,n)$ is when the latter consists of a single point, i.e. when $k = 0$ or $k = n$.

It then seems to be that I would need all six affine sets $U_{i,j}$ to completely cover the the grassmannian, is this true? 

Yes, in the sense that such a covering is minimal, since $[e_i \wedge e_j] \in G(2,n)$ is in $U_{ij}$ but not in any of the other $U_{k\ell}$.
Of course this doesn't rule out that there's some other open affine cover including some sets not of the form $U_{ij}$ that includes fewer open affines.  In fact, see https://mathoverflow.net/a/13482/6427

And finally, did I understand everything properly? Is it true to say that these are the open sets that make the grassmannian into a manifold of dimension $k(n-k)$?

Well, any open cover of a manifold by simply-connected sets gives you an atlas of the manifold.  So, yes, this one in particular will do.  If you didn't already know it was a manifold, you could prove it by taking these sets and exhibiting the relevant transition functions.
