# 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.)

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Show that any two bases can be reached from each other by row operations and look at what row operations do to the volume. –  Qiaochu Yuan Jan 16 '11 at 2:04
Adding one row to another does not change the volume, while multiplying a row by a scalar changes the volume by that scalar factor. So if we go from $e_i$'s to another basis, say $f_i's$, then the ratio $vol(f_1, \ldots, f_n)/vol(e_1, \ldots, e_n)$ is the product of all the scalar multiplications. But I don't see why this is the same as $vol(Tf_1, \ldots, Tf_n)/vol(Te_1, \ldots, Te_n)$ –  Alan C Jan 16 '11 at 2:21
You are half way there. The point is that since $T$ is linear, it commutes with scalar multiplication. So if $f_1=\alpha e_1$ for a scalar $\alpha$, then $Tf_1 = \alpha Te_1$. –  Alex B. Jan 16 '11 at 5:10
Aha thanks! I wonder how I overlooked the linearity of $T$. Hm... but I'm also wondering if this could be done in a way similar to the Steinitz exchange lemma. Suppose we order the $e_i$'s and $f_i$'s so that for all $0 \leq k \leq n$, $f_1, \ldots, f_k, e_{k+1}, \ldots, e_n$ span $V$. Then we could go from $vol(Te_1, \ldots, Te_n)/vol(e_1, \ldots, e_n)$ to $vol(Tf_1, \ldots, Tf_n)/vol(f_1, \ldots, f_n)$, replacing one vector at a time. (And the reason the ratio stays the same is because $T$ is linearity.) –  Alan C Jan 16 '11 at 15:28
In light of my comments on a previous, closely related question: in case you were wondering, I am not teaching linear algebra this semester! –  Pete L. Clark Feb 15 '11 at 16:48

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.

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