Plücker embedding - Two definitions While reading several papers on the topic of the Grassmannian, I cam about two definitions of the Plücker embedding.
One given as
$$ \varphi: \mathbb{A}^{n \cdot d} \rightarrow \mathbb{P}^{\binom{n}{d}-1}, A \mapsto \text{det}(A^{(I)})$$
where $ A^{(I)}$ denotes the submatrix of $A$ obtained by choosing its $I$-th columns. (I know that I can identify the affine space of this dimension with the space of matrices of size $d \times n$. Moreover, the map is  only well-defined if we look at matrices of rank $d$ which gives us the connection to elements of the Grassmannian). The other one was given as
$$\psi: Gr(d,n) \rightarrow \mathbb{P}^{\binom{n}{d}-1}, U \mapsto [u_1 \wedge \dotsc \wedge u_d]$$
So my question concerns the connection of these two maps in the projective space respectively whether there is a connection at all. I assume there must be something as in the papers I've read different authors used them both for talking about the Plücker embedding.
Thank you for your help.
 A: The idea here is that that the vectors $u_i, i=1,2,\ldots,d,$ are the rows of the matrix $A$. More precisely, if $u_i=(a_{i1},a_{i2},\ldots,a_{in})$, then the coefficient of 
$e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_d}$ in the wedge product $u_1\wedge u_2\wedge\cdots\wedge u_d$ is equal to $\det(A^{(I)})$.
Consider the paper & pencil example of $n=4,d=2$ with $u_1=(a_1,a_2,a_3,a_4)$, $u_2=(b_1,b_2,b_3,b_4)$ when
$$
\begin{aligned}
u_1\wedge u_2&=(a_1e_1+a_2e_2+a_3e_3+a_4e_4)\wedge(b_1e_1+b_2e_2+b_3e_3+b_4e_4)\\
&=\sum_{i=1}^4\sum_{j=1}^4a_ib_je_i\wedge e_j\\
&=(a_1b_2-a_2b_1)e_1\wedge e_2+(a_1b_3-a_3b_1)e_1\wedge e_3+(a_1b_4-a_4b_1)e_1\wedge e_4\\
&+(a_2b_3-a_3b_2)e_2\wedge e_3+(a_2b_4-a_4b_2)e_2\wedge e_4+(a_3b_4-a_4b_3)e_3\wedge e_4.
\end{aligned}
$$
See the six $2\times2$ minors of the matrix
$$
A=\left(\begin{array}{cccc}a_1&a_2&a_3&a_4\\b_1&b_2&b_3&b_4\end{array}\right)
$$
emerging?!

For the purposes of introducing coordinates to the Grasmannian we associate a $d$-dimensional subspace $V$ with any matrix $d\times n$ matrix $A$ that has $V$
as its row space. It is essential that:


*

*Two matrices $A$ and $A'$ are  sharing the same row space if and only if there exists an invertible $d\times d$ matrix $M$ such that $A'=MA$.

*When we calculate the wedge products of the rows, all the coordinates of the product of the rows of $A'$ are gotten by multiplying the corresponding coordinates of the product of rows of $A$ by $\det M$. Implying that as homogeneous coordinates in a projective space the two wedge products refer to the same point. This means that the mapping $G(n,d)\to\Bbb{P}^N$ is well defined.

*The easiest way of proving the result of the previous bullet is to prove it by direct observation for all three types of elementary matrices, and then writing $M$ as a product of elementary matrices.


Also, proving that $\phi$ and $\psi$ really give the same mapping may be easiest to do by induction on $d$. Then, at the induction step, you get the usual expansions of $(d+1)\times (d+1)$-determinants as linear combinations of $d\times d$-determinants.
