Exterior power of multilinear functions applied to linearly dependent vectors is zero

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 $V \times V$.

migrated from mathoverflow.netMar 12 '15 at 21:27

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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.
• @Samadwara Reddy I guess, if you learned exterior forms in some algebra course or something, it is customary to define tensor and exterior products via free vector spaces and quotient spaces, and I guess that obscures this. Basically, because $V$ is finite-dimensional, it is reflexive ($V$ can be identified with $V^{**}$, you can view the duality relation between $V$ and $V^*$ as completely symmetric, so "ordinary" vectors are also functionals on dual vectors. With this in mind, wedge products are a way of creating alternating multilinear functionals. – Bence Racskó Mar 13 '15 at 11:31
• @SamadwaraReddy Also, I don't really understand your last statement, but if $e^1,...,e^n$ is a basis for $V^*$, then $T$ can be written as $$T=\sum_{\mu_1<...<\mu_p}T_{\mu_1,...\mu_p}e^{\mu_1}\wedge...\wedge e^{\mu_p}.$$ We call an element of an exterior product space "simple" if it can be written as a wedge product without any sums. – Bence Racskó Mar 13 '15 at 11:35