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.net Mar 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.

  • $\begingroup$ Thanks for answering! One last thing. The reason I hadn't already solved the problem was that I thought the alternating property wasn't applicable in this case. I thought it applied to wedge products of vectors that repeat. But here we are looking at wedge products of linear functions of repeated vectors, right? (Also, about the simple exterior form thing for T, would the correct statement be that it could be written as an element in the span of vectors of that form?) $\endgroup$ – Nezo Mar 12 '15 at 23:27
  • $\begingroup$ @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. $\endgroup$ – Bence Racskó Mar 13 '15 at 11:31
  • $\begingroup$ @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. $\endgroup$ – Bence Racskó Mar 13 '15 at 11:35

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