Why are Alternating Forms Zero when Evaluated for a Set of Linear Dependent Vectors?

Question

I am reading a textbook on differential geometry and gauge theories and it it says off-hand that the alternating property of alternating multilinear forms implies that a form of this kind evaluated on a linearly dependent set of vectors is zero.

Would someone mind clarifying why the alternating property implies this? For example, let's say it's a 2-linear form which maps a pair of vectors in the same vector space to the reals, we switch the order of the pair in the Cartesian product and that changes the sign of the real which we map to.

$\phi(v_2,v_1)=-\phi(v_1,v_2)$

If the two vectors are linearly dependent so there exists a weighted combination of them added together which adds to zero (or equivalent definition), why does that imply that evaluating the 2-form on that pair of vectors has to equal 0? I have studied differential forms and other exterior algebra, so any explanation in terms of these is fine, I would just like to know the proof for this.

EDIT: I did specify the multilinear case but to I will see if I can generalize to the trilinear case to answer the comment below, and then you just keep adding on indefinitely for a general case (can someone correct if this is wrong).

$\phi(v_1,v_2,v_3)=-\phi(v_3,v_2,v_1)$

If any of these vectors equals 0 the form is 0 by multilinearity. From linear dependence, a vector in the set can be expressed in terms of the others, say

$v_3 = av_2 + bv_2$

$\phi(v_1,v_2,av_1 + bv_2)=-\phi(av_1 + bv_2,v_2,v_1)$

$\phi$ is trilinear so

$\phi(v_1,v_2,av_1 + bv_2) = a\phi(v_1,v_2,v_1) + b\phi(v_1,v_2,v_2)$

$-\phi(av_1 + bv_2, v_2, v_1)=-a\phi(v_1,v_2,v_1)-b\phi(v_1,v_2,v_2)$

$a\phi(v_1,v_2,v_1) + b\phi(v_1,v_2,v_2)=-a\phi(v_1,v_2,v_1)-b\phi(v_1,v_2,v_2)$

This implies that $\phi(v_1,v_2,v_3)=0$.

For the $k$th case:

$\phi(v_1,...,v_{k-1},v_k)=-\phi(v_k,v_{k-1},...,v_1)$

From linear dependence:

$v_k=av_1+...+dv_{k-1}$

$\phi(v_1,...,v_{k-1},av_1 +...+dv_{k-1})=-\phi(av_1+...+dv_{k-1},...,v_1)$

$\phi$ is $k$-linear so

$\phi(v_1,...,v_{k-1},av_1+...+dv_{k-1})=a\phi(v_1,...,v_{k-1},v_1)+...+d\phi(v_1,...,v_{k-1},v_{k-1})$

$-\phi(av_1...+dv_{k-1},v_{k-1},...,v_1)=-a\phi(v_1,v_{k-1},...,v_1)-...-d\phi(v_{k-1},v_{k-1},...,v_1)$

$a\phi(v_1,...,v_{k-1},v_1)+...+d\phi(v_1,...,v_{k-1},v_{k-1})=-a\phi(v_1,v_{k-1},...,v_1)-...-d\phi(v_{k-1},v_{k-1},...,v_1)$

Hence $\phi(v_1,...,v_{k-1},v_k)=0$.

Answer 1

If either of the vectors are zero, then by bilinearity, ϕ($v_2$,$v_1$) = 0.

If $v_1$ and $v_2$ are linearly dependent nonzero vectors, then there must be some scalar c such that $v_1 = cv_2$. Then the statement

ϕ($v_2$,$v_1$)=−ϕ($v_1$,$v_2$)

can be rewritten

ϕ($v_2$,$cv_2$)=−ϕ($cv_2$,$v_2$)

But if ϕ is bilinear, then ϕ($v_2$,$cv_2$) = cϕ($v_2$,$v_2$) and ϕ($cv_2$,$v_2$) = cϕ($v_2$,$v_2$)

So we have

cϕ($v_2$,$v_2$) = -cϕ($v_2$,$v_2$)

Thus, ϕ($v_2$,$v_2$) = 0

Answer 2

For any vector $v$, $$\phi(v,v)=-\phi(v,v)$$ Therefore (assuming the base field has characteristic different from $2$) $$\phi(v,v)=0.$$

Now if $v_2=\lambda v_1$, by bilinearity $$\phi(v_1,v_2)=\lambda\phi(v_1,v_1)=0.$$