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Let $V$ denote a vector space (maybe not finite-dimensional) over a field $\mathbb k$ with basis $\{e_1, e_2, \ldots\}$. I have to prove that the set $\{e_{i_1} \wedge e_{i_2} \ldots \wedge e_{i_n}$ $|$ $i_j < i_{j+1}$, $n \in \mathbb N$, $n \leq \dim V\}$ is a basis in $\bigwedge V$.

Of course, those vectors spans the exterior algebra. My problem is their linear independence.

Initially I said that we can talk only about a finite-dimensional spaces (we can obviously show that the infinite-dimensional case is almost similar to finite-dimensional).

After that we can take a zero linear combination of the pure tensors. We can say that there is some $j$ such that some tensors contain $e_j$ and some not. Let's take an exterior product of our combination and $e_j$. This new combination will contain less number of pure tensors, but it's still zero.

So, by induction, now our goal is to prove that for different $e_{i_j}$ $e_{i_1} \wedge e_{i_2} \ldots \wedge e_{i_n} \neq 0$.

How to do it?

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You have the idea, take a linear combination $c=\sum a_{i_1...i_k}e_{i_1}...\wedge e_{i_k}$ and take the wedge product of $c$ with $e_{j_1...j_{n-k}}$ where $(e_{i_1},..,e_{i_k},e_{j_1},...,e_{j_{n-k}})$ is a basis. The result is $a_{i_1..i_k}e_{i_1}\wedge..\wedge e_{i_k}\wedge e_{j_1}\wedge...\wedge e_{j_{n-k}}$.

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  • $\begingroup$ Yes, I've done this by myself. But how to prove that your result $a_{i_1..i_k}e_{i_1}\wedge..\wedge e_{i_k}\wedge e_{j_1}\wedge...\wedge e_{j_{n-k}}$ isn't a zero? $\endgroup$ – Vremennik Dec 29 '18 at 14:15
  • $\begingroup$ Because you have only one term in the result of $c\wedge e_{j_1...j_{n-k}}$ which is $a_{i_1..i_k}e_{i_1}\wedge..\wedge e_{i_k}\wedge e_{j_1}\wedge...\wedge e_{j_{n-k}}=0$ so $a_{i_1...i_k}=0$. $\endgroup$ – Tsemo Aristide Dec 29 '18 at 14:17
  • $\begingroup$ But you don't say anything about the case if $e_{i_1}\wedge..\wedge e_{i_k}\wedge e_{j_1}\wedge...\wedge e_{j_{n-k}} = 0$. Why it's unreal? $\endgroup$ – Vremennik Dec 29 '18 at 14:19
  • $\begingroup$ because $(e_{i_1},...,e_{i_k},e_{j_1},...,e_{j_{n-k}})$ is a basis. $\endgroup$ – Tsemo Aristide Dec 29 '18 at 14:20
  • $\begingroup$ And how my statement implies from this? $\endgroup$ – Vremennik Dec 29 '18 at 14:21

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