# The non-zero exterior product

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?

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 $$a_{i_1..i_k}e_{i_1}\wedge..\wedge e_{i_k}\wedge e_{j_1}\wedge...\wedge e_{j_{n-k}}$$.
• 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? – Vremennik Dec 29 '18 at 14:15
• 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 $a_{i_1...i_k}=0$. – Tsemo Aristide Dec 29 '18 at 14:17
• 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? – Vremennik Dec 29 '18 at 14:19
• because $(e_{i_1},...,e_{i_k},e_{j_1},...,e_{j_{n-k}})$ is a basis. – Tsemo Aristide Dec 29 '18 at 14:20