algebraic manipulation of differential form Suppose $\phi_1, \phi_2, \dots, \phi_k \in (\mathbb{R}^n)^*$, and $\mathbf{v}_1, \dots, \mathbf{v}_k \in \mathbb{R}^n$ 
$(\mathbb{R}^n)^*$ stands for the space of all linear transformations that goes from $\mathbb{R}^n \to \mathbb{R}$
is it true that:
$$\phi_1\wedge\dots\wedge\phi_k(\mathbf{v}_1, \dots, \mathbf{v}_k) = \mathrm{det}[\phi_i(\mathbf{v}_j)]$$
This is a homework question (from Multivariable Mathematics, Shifrin, Ex.18 on Page 347) and there is a hint that:we could first express all $\phi_i$ in the form of:
$$\phi_i = \sum\limits_{j=1}^n a_{ij}dx_j$$
where $dx_j$ is the basis 1-form
and the hint also says that it suffices to prove this equality in the case where all $v_j$ are standard basis vectors
i.e. $v_1=e_{j_1}, \dots, v_k = e_{j_k}$
From the setup of the hint, it is quite obvious that the right side equals:
$$
\begin{vmatrix}
a_{1j_1}&\cdots&a_{1j_k}\\
\vdots&\ddots&\vdots\\
a_{kj_1}&\cdots&a_{kj_k}\\
\end{vmatrix}
$$
However, I cannot understand how to show the left side equals this determinant.
Also, I am not very sure why it suffices to show that the equality holds when $v_j$ are all standard basis vectors. Is it because I could express every $v_j$ as a linear combination of the standard basis vectors?
Thank you very much!
 A: Consider the following alternation operator:
$$
\operatorname{Alt}(P) = \sum_{\sigma \in S_n} (-1)^{sgn(\sigma)}P(x_{\sigma(1)}, \dots, x_{\sigma(n)})
$$
where $P(x_1,\dots,x_n)$ is a polynomial of $n$ variables.
Let us regard a matrix $A=(a_{i j})_{i,j=1,\dots,k}$ as a collection of its columns:
$$
A = (A_1, \dots, A_k)
$$
Operators $dx_i$ act on these columns as
$$
dx_i(A_j) := a_{i j}
$$
The determinant of $A$ may be defined as
$$
\det(A) := \operatorname{Alt}(dx_1(A_1)\dots dx_k(A_k))
$$
where the alternation is applied to the polynomial in variables $dx_i$ (in other words, the order of $A_j$'s is kept unchanged when alternating).
The wedge product's definition is
$$
\phi_1 \wedge \dots \wedge \phi_k := \operatorname{Alt}(\phi_1 \otimes \dots \otimes \phi_k)
$$
Substituting the data from the question shows that 
$$
\phi_1\wedge\dots\wedge\phi_k(\mathbf{v}_1, \dots, \mathbf{v}_k) = \mathrm{det}[\phi_i(\mathbf{v}_j)]
$$
holds tautologically for $\phi_i = dx_i$.
A: *

*we have to show that $\phi_1\wedge \dots \wedge \phi_k(v_1, \dots, v_k) = \mathrm{det}[\phi_i(v_j)]$ is equivalent to
$\phi_1\wedge \dots \wedge \phi_k(e_{j_1}, \dots, e_{j_k}) = \mathrm{det}[\phi_i(e_{j_l})]$, $i = 1, \dots, k$ and $l = 1, \dots k$
Suppose $\phi_1\wedge \dots \wedge \phi_k(v_1, \dots, v_k) = \mathrm{det}[\phi_i(v_j)]$, when the $v_j$ are not linearly independent, we have both side equal to zero.
When the $v_j$ are not linearly independent, these k vectors span a k-dimensional subspace of $\mathbb{R}^n$, let $e_{j_1}, \dots,e_{j_k}$ be the set of vectors that also span this k-dimensional subspace.
Hence, each $e_{j_l}$ could be expressed as a linear combination of the $v_j$ vectors.
$e_{j_1} = c_{11}v_1 + \dots + c_{1k}v_k$
$e_{j_2} = c_{21}v_1 + \dots + c_{2k}v_k$
$\vdots$
$e_{j_k} = c_{k1}v_1 + \dots + c_{kk}v_k$
Since k-forms and determinants are both multilinear,
We perform these linear combinations on both sides of the original equation.
$\phi_1\wedge \dots \wedge \phi_k(c_{11}v_1 + \dots + c_{1k}v_k, \dots, c_{k1}v_1 + \dots + c_{kk}v_k) = det[\phi_i(c_{l1}v_1 + \dots + c_{lk}v_k)]$
$\phi_1\wedge \dots \wedge \phi_k(e_{j_1}, \dots, e_{j_k}) = det[\phi_i(e_{j_l})]$
Hence, we get the conclusion that the original equality is equivalent to the equality we get by substituting the proper standard basis vectors.

*Now we have to explain why the equality holds.
on the right hand side, we have:
$$\begin{vmatrix}
 \phi_1(e_{j_1})&\dots&\phi_1(e_{j_k})\\
 \vdots&\ddots&\vdots\\
 \phi_k(e_{j_1})&\dots&\phi_k(e_{j_k})
\end{vmatrix}$$
for each $\phi_i(e_{j_l})$, we have $\phi_i = \sum\limits_{j = 1}^n a_{ij}dx_{j}$
Since we know that $dx$ is a linear 1-form, then $\phi_i(e_{j_l})$ is equal to the sum of each $dx_j$ applied to $e_{j_l}$.
Because $e_{j_l}$ is a standard basis vector, there will be only one term that does not produce zero, which is $a_{ij_l}dx_{j_l}$.
Hence, $\phi_i(e_{j_l}) = 0+\dots+0 + a_{ij_l}dx_{j_l}(e_{j_l})+0+\dots+0 = a_{ij_l}$
Hence, the determinant on the right hand side becomes
$$\begin{vmatrix}
 a_{1j_1}&\dots&a_{1j_k}\\
 \vdots&\ddots&\vdots\\
 a_{kj_1}&\dots&a_{kj_k}
\end{vmatrix}$$
Now let's examine the left hand side:
Express each $\phi_i = \sum\limits_{j = 1}^n a_{ij}dx_j$
Hence,
$\phi_1\wedge \dots \wedge \phi_k = \sum\limits_{\text{all possible }I} C_I dx_I$, where $I$ is a k-tuple of natural numbers from 1 to n.
if we apply this k-form to $e_{j_1}, \dots, e_{j_k}$, there will be exactly one $I$ which produces a non-zero number, and that $I$ is equal to $\{j_1, j_2, \dots, j_k\}$
Hence, the only term that produces a non-zero results is the term:
$C_{j_1, \dots, j_k}dx_{j_1}\dots dx_{j_k}(e_{j_1},\dots, e_{j_k}) = C_{j_1, \dots, j_k}$
Now we simply need to figure out what $C_{j_1, \dots, j_k}$ is.
We realize that $C_{j_1, \dots, j_k}$ comes from the product of the coefficients of all perumtations of $\{dx_{j_1}, \dots, dx_{j_k}\}$. Thus,
$C_{j_1, \dots, j_k} = \sum\limits_{\text{all permutations of }\{j_1, \dots, j_k\}} sgn(\sigma)\prod_{i = 1}^k a_{i\sigma(j_i)}$,
which is the determinant:
$$
 \begin{vmatrix}
  a_{1j_1}&\dots&a_{1j_k}\\
  \vdots&\ddots&\vdots\\
  a_{kj_1}&\dots&a_{kj_k}
 \end{vmatrix}
$$
And this is equal to the right hand side.
Hence, the original statement is proven.
