Question about the proof of a theorem in Differential geometry I'm currently reading Michael Spivak's "Calculus on Manifolds" since I need to learn some basics for a Electrodynamics class. The question is about Theorem 4-6, or more precisely about the proof of it.
Theorem
Let $\{v_i\}_{i=1}^n$  be a basis for $V$, and let $\omega \in \Lambda^n(V)$. If $w_i=\sum_{j=1}^n a_{ij}v_j$ are $n$ vectors in $V$, then 
$$\omega(w_1,\dots, w_n)=\operatorname{det}(a_{ij})\cdot \omega(v_1,\dots,v_n).$$
Proof
Define $\eta\in\mathfrak{I}^n(\mathbb{R}^n)$ by 
$$\eta((a_{11},\dots,a_{1n}),\dots,(a_{n1},\dots,a_{nn}))=\omega\left(\sum a_{1j}v_j,\dots,\sum a_{nj}v_j\right).$$
$\eta\in\Lambda^n(\mathbb{R}^n)$ so $\eta =\lambda \cdot \operatorname{det}$ for some $\lambda\in\mathbb{R}$ and $\lambda=\eta(e_1,\dots,e_n)=\omega(v_1,\dots,v_n)$.
$\hspace{16cm}\Box$
I really don't understand any part of this proof, so it would be great if someone could explain this in a little bit more detail. Please keep in mind that I'm still not very familiar with the introduced concepts.
Edit: $\mathfrak{I}^n(V)$ is the set of all $n$-Tensors on $V$.
 A: I assume $V$ is a vector space over $\mathbb R$. I wonder if the notation would be more clear if we introduce the linear transformation $T:\mathbb R^n \to V$ defined by 
$$
T\left(\begin{bmatrix} c_1 \\ \vdots \\ c_n \end{bmatrix} \right)
= c_1 v_1 + \cdots + c_n v_n.
$$
If $a_1,\ldots,a_n \in \mathbb R^n$, then
$$
\eta(a_1,\ldots,a_n) = \omega(T(a_1),\ldots,T(a_n)).
$$
The proof asserts that $\eta$ is multilinear, but let's check this more carefully:
\begin{align}
\eta(a_1 + a_1',a_2,\ldots,a_n) &= \omega(T(a_1 + a_1'),T(a_2),\ldots,T(a_n)) \\
&= \omega(T(a_1) + T(a_1'),T(a_2),\ldots,T(a_n)) \\
&= \omega(T(a_1),T(a_2),\ldots,T(a_n)) + \omega(T(a_1'),T(a_2),\ldots,T(a_n)) \\
&= \eta(a_1,a_2,\ldots,a_n) + \eta(a_1',a_2,\ldots,a_n).
\end{align}
(In the second-to-last step we used the fact that $\omega$ is multilinear.)
Also,
\begin{align}
\eta(\alpha a_1,a_2,\ldots,a_n) &= \omega(T(\alpha a_1),T(a_2),\ldots,T(a_n)) \\
&= \omega(\alpha T(a_1),T(a_2),\ldots,T(a_n)) \\
&= \alpha \omega(T(a_1),T(a_2),\ldots,T(a_n)) \\
&= \alpha \eta(a_1,a_2,\ldots,a_n).
\end{align}
This shows that $\eta$ is linear in its first argument. Similarly, we can show that $\eta$ is linear in each of its arguments.
The proof also asserts that $\eta$ is alternating, so let's check that too.
\begin{align}
\eta(a_2,a_1,a_3,\ldots,a_n) &= \omega(T(a_2),T(a_1),T(a_3),\ldots,T(a_n)) \\
&=-\omega(T(a_1),T(a_2),\ldots,T(a_n)) \\
&= -\eta(a_1,a_2,\ldots,a_n).
\end{align}
This shows that interchanging the first two inputs to $\eta$ flips the sign of the output. Similarly, we can show that interchanging the $i$th and $j$th inputs to $\eta$ flips the sign of the output.
So, we have shown in more detail that $\eta \in \Lambda^n(\mathbb R^n)$.
But, the book has shown previously that $\Lambda^n(\mathbb R^n)$ is one-dimensional. The most famous element of $\Lambda^n(\mathbb R^n)$, of course, is the determinant function $\det$. It follows that
$$
\tag{$\spadesuit$} \eta = \lambda \det
$$
for some scalar $\lambda$.
And what is $\lambda$? Let $e_1,\ldots,e_n$ be the standard basis vectors for $\mathbb R^n$. We can find the value of $\lambda$ by plugging the inputs $e_1,\ldots,e_n$ into both sides of equation ($\spadesuit$):
\begin{align}
\eta(e_1,\ldots,e_n) &= \lambda \det(e_1,\ldots, e_n) \\
&= \lambda.
\end{align}
But notice also that
\begin{align}
\eta(e_1,\ldots,e_n) &= \omega(T(e_1),\ldots, T(e_n) ) \\
&= \omega(v_1,\ldots,v_n).
\end{align}
This shows that $\lambda = \omega(v_1,\ldots,v_n)$.
Thus, from equation ($\spadesuit$), we have
$$
\eta(a_1,\ldots,a_n) = \omega(v_1,\ldots,v_n) \det(a_1,\ldots,a_n).
$$
In other words,
$$
\omega(T(a_1),\ldots,T(a_n)) = \omega(v_1,\ldots,v_n) \det(a_1,\ldots,a_n).
$$
This is what we wanted to show.
