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The wikipedia page on exterior algebras makes the following reasonable sounding statement (I paraphrase):

Let $V$ be a complex vector space and consider the second exterior power $\bigwedge^2 V$. By the rank of $\alpha \in V$, we mean the smallest number $p$ such that $\alpha$ can be written as a sum of $p$ decomposable elements. CLAIM: $\alpha$ has rank $p$ if and only if the $p$-fold wedge product $\alpha \wedge \dots \wedge \alpha$ is nonzero but the $(p+1)$-fold wedge product is 0.

Unfortunately I can't figure out a proof of this, even in the case $p=1$, though it feels like it ought to be elementary. I've looked in a bunch of algebra books, but none of them explain it, probably because it is really easy and I am just missing something.

Please help!

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Suppose $\alpha$ can be written as a sum of $p$ decomposable elements, so that $$\alpha=v_1\wedge w_1+\cdots+ v_p\wedge w_p.$$

Let $q\geq1$. The $q$th power of $\alpha$ is then $$\alpha^q = \sum_{i_1,\dots,i_q} (v_{i_1}\wedge w_{i_1})\wedge(v_{i_2}\wedge w_{i_2})\wedge\cdots\wedge (v_{i_q}\wedge w_{i_q})$$ where the sum is taken over all choices of indices $i_1$, $i_2$, $\dots$, $i_q$ in $\{1,\dots,q\}$.

Consider a term in this sum, corresponding to a choice of $i_1$, $i_2$, $\dots$, $i_q$. If two of the $i_j$s are equal, the term vanishes. If $q\leq p$; if the $i_j$s are all different, then since each $v_{i_j}\wedge w_{i_j}$ has degree two, we see than the term is equal to $(v_1\wedge w_1)\wedge(v_2\wedge w_2)\wedge\cdots(v_p\wedge w_q)$.

It follows from this that if $q>p$ then $\alpha^q=0$, for then all terms vanish, and then the rank of $\alpha$ is an upper bound for the highest non-zero power of $\alpha$.

On the other hand, if $q=p$, then there are $p!$ ways of choosing the indices $i_1$, $i_2$, $\dots$, $i_p$ in $\{1,\dots,p\}$ so that they are distinct, so we see that $$\alpha^p = p!\cdot (v_1\wedge w_1)\wedge(v_2\wedge w_2)\wedge\cdots(v_p\wedge w_p).$$

Can you see why $\alpha^{\mathrm{rank}\alpha}\neq0$?

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    $\begingroup$ I don't understand your statement "If $q\leq p$; if the $i_j$s are all different, then since each $v_{i_j}\wedge w_{i_j}$ has degree two, we see than the term is equal to $(v_1\wedge w_1)\wedge(v_2\wedge w_2)\wedge\cdots(v_p\wedge w_q)$." Surely the term is only equal to some $(v_{i_1}\wedge w_{i_1})\wedge(v_{i_2}\wedge w_{i_2})\wedge\cdots\wedge (v_{i_q}\wedge w_{i_q})$ where $i_1, \dots, i_q$ is a strictly increasing sequence in $\{1, \dots, p \}$? How can you assume that the sequence is $1, 2, \dots, q$? $\endgroup$ – Oliver Apr 11 '11 at 2:32
  • $\begingroup$ @Oliver: I had a mixture of two things... Hopefully now it makes sense. $\endgroup$ – Mariano Suárez-Álvarez Apr 11 '11 at 3:59

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