Defining the determinant of linear transformations as multilinear alternating form Here is what our professor showed us in our linear algebra class to introduce the idea of determinants:
Suppose we have an $n$-dimensional vector space $V$. Then we can create a function from $V^n$ to $\mathbb{R}$ called $vol$ (for "volume") satisfying these properties:
$vol$ is multilinear
$vol$ is alternating (i.e. if any two of $v_1, \ldots, v_n$ are the same, then $vol(v_1, \ldots, v_n) = 0$)
From these two properties, we can see that if $e_1, \ldots, e_n$ is a basis of $V$, then the $vol$ function is completely defined by the value $vol(e_1, \ldots, e_n)$. 
Thus if $T$ is a linear operator on $V$, the ratio:
$\dfrac{vol(Te_1, \ldots, Te_n)}{vol(e_1, \ldots, e_n)}$
is the same for any (multilinear and alternating) $vol$ function.
However, I am having trouble understanding why the ratio is also independent of the basis $e_1, \ldots, e_n$. This is what I am asking for help with. I can see that this invariance implies that intuitively, every $n$-parallelotope is stretched by the same amount by the operator $T$.
(Our professor then defined the determinant of $T$ as that ratio.)
 A: First note that since $T$ is linear,  $\mathrm{vol}_T(v_1, \dots, v_n) = \mathrm{vol} (T v_1 , \dots , T v_n )$ is also a multilinear alternating function. 
So if you go from $e_1, \dots , e_n$ to a different basis, say, $b_1, \dots b_n$, then you can write the $b_i$ in terms of their coordinates with respect to $e_1, \dots , e_n$:
$$
 b_1= \sum_{i=1}^n b_{i  1} \, e_i,\quad  b_2= \sum_{i=1}^n b_{i  2} \, e_i,\quad\dots \quad b_n= \sum_{i=1}^n b_{i  n} \, e_i.
$$
And Leibniz's formula (which is a direct consequence of multinearity and alternating-ness), applied to both $\mathrm{vol}$ and $\mathrm{vol}_T$ tells us:
$$\mathrm{vol}(b_1 , \dots , b_n ) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \: b_{\sigma(1) 1} \cdots b_{\sigma(n) n} \cdot \mathrm{vol}(e_1, \dots , e_n)$$
$$\mathrm{vol} (T b_1 , \dots , T b_n ) =  \sum_{\sigma \in S_n} \text{sgn}(\sigma) \: b_{\sigma(1) 1} \cdots b_{\sigma(n) n} \cdot \mathrm{vol}(T e_1, \dots , T e_n)$$
So if you take the ratio $\displaystyle \frac{\mathrm{vol} (T b_1 , \dots , T b_n )}{\mathrm{vol} (b_1 , \dots , b_n )} $ everything else cancels out and you're left with $ \displaystyle \frac{\mathrm{vol} (T e_1 , \dots , T e_n )}{\mathrm{vol} (e_1 , \dots , e_n )}$ again.
Of course, for all of this, it is important that $\mathrm{vol} \neq 0$. That's probably an additional restriction your professor put on the $\mathrm{vol}$ function.
A: Let $K$ be the field of the scalar multiplication of V.
Define $\Lambda_n(V)$ as the set of all alternating n-forms of V.
It is easy to prove that $\Lambda_n(V)$ is a $K$-vector space if equipped with the sum of vectors and multiplication by a scalar defined for every $f,g\in\Lambda_n(V)$ and every $k\in K$ respectively by
$$(f+g)(v_1, ...,v_n)=f(v_1,...,v_n)+g(v_1,...v_n)$$
$$(kf)(v_1,...,v_n)=kf(v_1,...,v_n)$$
$v_1,...v_n\in K$.
The ratio you defined turns out to depend only on the operator $T$ from the fact that $\Lambda_n(V)$ as dimension 1: if $\phi,\psi\in\Lambda_n(V)$ then there exists a $\lambda\in K$ such that $\phi=\lambda\psi$.
