Understanding the properties of multilinear alternating maps

Question

I recently came across this proposition in my math textbook:

Let $D: \mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}^3 \rightarrow \mathbb{R}$ be a multilinear, alternating map such that $D(a_1,a_2,a_3)=1$. Let $v_1,v_2,v_3 \in V$. If $D(v_1,v_2,v_3) = 0$ then $span(v_1,v_2,v_3) \neq \mathbb{R}^3$.

It is strange to me how my linear algebra textbook can make such a bold claim without an accompanying proof or explanation. I would greatly appreciate it if someone could help me understand what the last statement is claiming exactly and if there is a proof available that can help me arrive at this statement. Thank you and any insights are greatly appreciated.

Answer 1

It's a well known result that the determinant can be defined as the unique $n$-alternating multilinear map which is $1$ at the basis $\{e_1, \cdots, e_n\}$ (see here). Your desired result follows immediately from this, for then the claim $D(v_1, v_2, v_3) = 0$ is equivalent to $v_1, v_2, v_3$ being linearly dependent, hence they can't generate all of $\mathbb{R}^3$.

Answer 2

You should try to understand Matheus's answer for the insight that it gives into the determinant. (It's surely no coincidence that your textbook used the first letter of ‘determinant’ for the name of this map!) But here is an outline of an elementary proof (following the idea of Reveillark's comment).

Suppose that $ \operatorname { span } ( v _ 1 , v _ 2 , v _ 3 ) = \mathbb R ^ 3 $. Since $ a _ 1 , a _ 2 , a _ 3 \in \mathbb R ^ 3 $, there exist $ 9 $ coefficients $ c _ 1 ^ 1 $, $ c _ 1 ^ 2 $, $ c _ 1 ^ 3 $, $ c _ 2 ^ 1 $, $ c _ 2 ^ 2 $, $ c _ 2 ^ 3 $, $ c _ 3 ^ 1 $, $ c _ 3 ^ 2 $, and $ c _ 3 ^ 3 $ such that $ a _ 1 = c _ 1 ^ 1 v _ 1 + c _ 1 ^ 2 v _ 2 + c _ 1 ^ 3 v _ 3 $, $ a _ 2 = c _ 2 ^ 1 v _ 1 + c _ 2 ^ 2 v _ 2 + c _ 2 ^ 3 v _ 3 $, and $ a _ 3 = c _ 3 ^ 1 v _ 1 + c _ 3 ^ 2 v _ 2 + c _ 3 ^ 3 v _ 3 $. Now since $ D $ is multilinear, $$ \eqalign { D ( a _ 1 , a _ 2 , a _ 3 ) & = D ( c _ 1 ^ 1 v _ 1 + c _ 1 ^ 2 v _ 2 + c _ 1 ^ 3 v _ 3 , c _ 2 ^ 1 v _ 1 + c _ 2 ^ 2 v _ 2 + c _ 2 ^ 3 v _ 3 , c _ 3 ^ 1 v _ 1 + c _ 3 ^ 2 v _ 2 + c _ 3 ^ 3 v _ 3 ) \\ & = c _ 1 ^ 1 c _ 2 ^ 1 c _ 3 ^ 1 D ( v _ 1 , v _ 1 , v _ 1 ) + c _ 1 ^ 1 c _ 2 ^ 1 c _ 3 ^ 2 D ( v _ 1 , v _ 1 , v _ 2 ) + c _ 1 ^ 1 c _ 2 ^ 1 c _ 3 ^ 3 D ( v _ 1 , v _ 1 , v _ 3 ) \\ & \qquad { } + c _ 1 ^ 1 c _ 2 ^ 2 c _ 3 ^ 1 D ( v _ 1 , v _ 2 , v _ 1 ) + c _ 1 ^ 1 c _ 2 ^ 2 c _ 3 ^ 2 D ( v _ 1 , v _ 2 , v _ 2 ) + c _ 1 ^ 1 c _ 2 ^ 2 c _ 3 ^ 3 D ( v _ 1 , v _ 2 , v _ 3 ) \\ & \qquad { } + c _ 1 ^ 1 c _ 2 ^ 3 c _ 3 ^ 1 D ( v _ 1 , v _ 3 , v _ 1 ) + c _ 1 ^ 1 c _ 2 ^ 3 c _ 3 ^ 2 D ( v _ 1 , v _ 3 , v _ 2 ) + \cdots + c _ 1 ^ 3 c _ 2 ^ 3 c _ 3 ^ 3 D ( v _ 3 , v _ 3 , v _ 3 ) \text . } $$ This consists of $ 3 ^ 3 = 2 7 $ terms, so I didn't write them all out, but I wrote out enough that you can see what they're like. The first $ 5 $ terms are all zero because $ D $ is alternating; in all, $ 2 1 $ terms are zero for this reason. The next term is zero because $ D ( v _ 1 , v _ 2 , v _ 3 ) = 0 $ is given. The next term is another one of the $ 2 1 $ terms that are zero because $ D $ is alternating. The last term that I wrote out before the ellipsis is zero because of a combination of these reasons: $ D ( v _ 1 , v _ 3 , v _ 2 ) = - D ( v _ 1 , v _ 2 , v _ 3 ) = - 0 = 0 $ (because an alternating multilinear map is antisymmetric, which I will assume that you've already seen).

So we see that $ D ( a _ 1 , a _ 2 , a _ 3 ) $ consists of $ 2 7 $ terms, all zero, so it must be zero, contradicting the given fact that it equals $ 1 $. The assumption that $ \operatorname { span } ( v _ 1 , v _ 2 , v _ 3 ) = \mathbb R ^ 3 $ is therefore false. (More directly, this proves that if $ D $ is multilinear and alternating and $ D ( v _ 1 , v _ 2 , v _ 3 ) = 0 $ for some triple $ ( v _ 1 , v _ 2 , v _ 3 ) $ of vectors, then $ D ( a _ 1 , a _ 2 , a _ 3 ) = 0 $ for every triple $ ( a _ 1 , a _ 2 , a _ 3 ) $ of vectors in $ \operatorname { span } ( v _ 1 , v _ 2 , v _ 3 ) $.)

By the way, if you start instead with $ \operatorname { span } ( a _ 1 , a _ 2 , a _ 3 ) = \mathbb R ^ 3 $ and take three arbitrary vectors $ w _ 1 $, $ w _ 2 $, and $ w _ 3 \in \mathbb R ^ 3 $, then you will prove what Matheus said, that $ D $ is the map that takes any three vectors to the determinant of the matrix whose rows (or columns) are those three vectors. Try it!