I've seen the following claim several times:

If $V$ is a vector space over $K$ with basis $\{e_1,\ldots,e_n\}$ then the basis to the kth exterior power of $V$ is given by the elements $$\{e_{i_1}\wedge e_{i_2}\wedge\cdots\wedge e_{i_k} \mid 1 \le i_1 < i_2 < \cdots < i_k \le n\}$$

Now, given some basic facts on exterior powers, it's not hard to show that the above set actually spans the $k$th exterior power. On the other hand, I've never seen a complete proof of the linear independence of this set. So my question is: What is the simplest way to show that $\{e_{i_1}\wedge e_{i_2}\wedge\cdots \wedge e_{i_k} \mid 1 \le i_1 < i_2 < \cdots < i_k \le n\}$ is linearly independent?

  • 3
    $\begingroup$ Careful; you use $k$ for both the underlying field and for an integer. $\endgroup$ – Qiaochu Yuan Jul 24 '11 at 23:29

We induct on $n - k$.

Suppose there exists a nontrivial linear dependence among the pure tensors you list in $\Lambda^k(V)$. Then not all of the tensors share the same components $e_i$, so there is some $i$ such that some tensor does not contain $e_i$ (and some other tensor does). Take the exterior product with $e_i$; this gives a linear dependence among a smaller number of pure tensors in $\Lambda^{k+1}(V)$. By repeating this argument we see that it suffices to show that no single pure tensor can be equal to zero. If $k < n$, then it suffices to take the exterior product with the rest of the basis so that we reduce to the case $k = n$. If $k = n$, it suffices to use any of the standard proofs that the determinant exists.


If you want an easy access reference, this book, which is available for free online has a proof of the statement you're looking for in section 2.3.2, basically it's their "lemma 3".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.