I'm working on a homework problem, and we are to show that if $T \in \wedge^p V^*$, and $v_1,\ldots,v_p$ are linearly dependent, then $T(v_1,\ldots,v_p) = 0$.
What I've got so far: I understand that we may write $T = t_1 \wedge t_2 \wedge \ldots \wedge t_n$ for $t_1,\ldots,t_n \in V*$ (multilinear functions). Additionally, by linear dependence, without loss of generality, we may write $v_1 = c_2 v_2 + \ldots + c_p v_p$.
I must now show that $(t_1 \wedge t_2 \wedge \ldots \wedge t_n)(c_2 v_2 + \ldots + c_p v_p,v_2,\ldots,v_p) = 0$. At this point, I got stuck, so I looked at the case of $p=2$, as seen below. (where $a_1,b_1,a_2,b_2$ are $n$-dimensional linear operators) $\begin{align*} T(v_1,v_2) &= T(v_1,cv_1) \\ &= t_1(v_1,cv_1) \wedge t_2(v_1,cv_1) \\ &= c(t_1(v_1,v_1) \wedge t_2(v_1,v_1))~(here~is~where~I~get~stuck?)\\ &= c(a_1v_1b_1v_1 \wedge a_2v_1b_1v_2) \\ &= c(a_1b_1|v_1|^2 \wedge a_2b_2|v_1|^2) \\ &= 0 \\ \end{align*} \\ $
(since the wedge product of linearly dependent vectors is 0).
Is this the correct approach? If not, where did I go wrong? Otherwise, could someone give me some direction as to how to generalize this?
EDIT: Actually, the above example doesn't make any sense at all because vectors in $V^*$ act on vectors in $V$ not $V \times V$.
You can only write $T$ as $$ T=t_1\wedge t_2\wedge...\wedge t_p $$ if $T$ is a simple exterior form, which might not be the case. But, assume that $$ \{v_1,...,v_p\} $$ is a dependent set, then there exists a nontrivial linear combination giving $$ \alpha_1v_1+...+\alpha_pv_p=0. $$ Now assume that $\alpha_i$ is nonzero (at least one of them is nonzero, since it is a nontrivial linear combination), in this case $$ v_i=-\frac{\alpha_1}{\alpha_i}v_1-...-\frac{\alpha_p}{\alpha_i}v_p, $$ where obviously $v_i$ is missing from the sum. Now write $$ T(v_1,...,v_i,...v_p)=T\left(v_1,...,-\frac{\alpha_1}{\alpha_i}v_1-...-\frac{\alpha_p}{\alpha_i}v_p,...,v_p\right), $$ and use alternating property to see that this is zero, since if you expand by linearity, you will have $(p-1)$ terms and all of them contain a repeated vector.
Edit: Sorry, I merely skimmed over your post, I now see that you basically got here as well, just couldn't go further, so I'll expand a little bit more. If you have my last expression, you can use linearity to turn $$ T\left(v_1,...,-\frac{\alpha_1}{\alpha_i}v_1-...-\frac{\alpha_p}{\alpha_i}v_p,...,v_p\right) $$ into $$ -\frac{\alpha_1}{\alpha_i}T\left(v_1,...,v_1,...,v_p\right)-\frac{\alpha_2}{\alpha_i}T\left(v_1,v_2,...,v_2,...,v_p\right)-...\mathrm{etc},$$ and as you can see, all of these terms will have one vector twice in the argument, and by the alternating property, all terms will vanish.