# Proof of proposition about exterior product

I am looking at the proof of the following proposition:

Let $\displaystyle{\phi, \ldots , \phi_k\in V^{\star}}$ and $\displaystyle{\psi_i=\sum_{j=1}^ka_{ij}\phi_j, \ i=1, \ldots , k}$ with a matrix $\displaystyle{(a_{ij})\in M(k\times k, \mathbb{R})}$. Then it holds that $\displaystyle{\psi_1\land \ldots \land \psi_k=\det (a_{ij})\cdot \phi_1\land \ldots \land \phi_k}$.

The proof is the following:

We see that $\displaystyle{\det (a_{ij})=\sum_{\pi\in \text{Per}_k}\text{sign}(\pi)a_{1\pi(1)}a_{2\pi(2)}\cdot \ldots \cdot a_{k\pi (k)}}$, where $\text{Per}_k$ is the group of all permutations of $1, \ldots , k$.

It holds that \begin{align*}\psi_2\land \ldots \land \psi_k&=\left (\sum_{j_1=1}^ka_{1j_1}\phi_{j_1}\right )\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ & = \sum_{j_1, \ldots j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k} \\ &= \sum_{\pi\in \text{Per}_k}^k\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)} \\ & = \sum_{\pi\in \text{Per}_k}^k\text{sign}(\pi)\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{1}\land \ldots \land \phi_{k} \\ & = \det (a_{ij})\phi_{1}\land \ldots \land \phi_{k}\end{align*} q.e.d.



I want to understand that proof.



By a property of the exterior product we have that $$\phi_1\land \ldots \land \left (\lambda \phi_i'+\mu\phi_i''\right )\land \ldots \land \phi_k=\lambda \phi_1\land \ldots \land \phi_i' \land \ldots \land \phi_k+\mu \phi_1\land \ldots \land \phi_i''\land \ldots \land \phi_k$$

By induction we can show that it holds that $$\phi_1\land \ldots \land \left (\sum_{j=1}^n\lambda_j \phi_{i_j}\right )\land \ldots \land \phi_k=\sum_{j=1}^n\lambda_j \phi_1\land \ldots \land \phi_{i_j} \land \ldots \land \phi_k$$ for each $n\geq 2$.



Therefore, at the first part of the proof we get \begin{align*}\psi_2\land \ldots \land \psi_k&=\left (\sum_{j_1=1}^ka_{1j_1}\phi_{j_1}\right )\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ & = \sum_{j_1=1}^k\left (a_{1j_1}\right )\phi_{j_1}\land \ldots \land \left (\sum_{j_k=1}^ka_{kj_k}\phi_{j_k}\right ) \\ &= \sum_{j_1, \ldots , j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k}\end{align*}



I haven't really understood the next step of the proof. Do we consider the permutation $\pi: (1, \ldots , k)\mapsto (j_1, \ldots , j_k)$ and so we get that $$\sum_{j_1, \ldots , j_k=1}^k\left (a_{1j_1}\ldots a_{kj_k}\right )\phi_{j_1}\land \ldots \land \phi_{j_k}=\sum_{\pi\in \text{Per}_k}\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}$$ ? But isn't $j_i$ the variable of the sum?

If we do that like that then we get that $$\sum_{\pi\in \text{Per}_k}\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\phi_{\pi (1)}\land \ldots \land \phi_{\pi (k)}=\sum_{\pi\in \text{Per}_k}\left (a_{1\pi(1)}\ldots a_{k\pi (k)}\right )\cdot \text{sign}(\pi)\phi_{1}\land \ldots \land \phi_{k}$$ right?

• The point of that step is: if two of the $j$'s are equal, say $j_s = j_t$ for $s \ne t$, then $\phi_{j_s}$ appears twice in the wedge product, so that term of the sum is equal to 0. – Daniel Schepler Sep 27 '17 at 22:34
• Ah ok. By the sum do we get all the possible tuples? And by using the permutation $\pi$ we get all the possible tuples with distict elements, or not? @DanielSchepler – Mary Star Sep 27 '17 at 22:38
• Exactly, the left hand side is the sum over all tuples, the right hand side is the equivalent sum over tuples where $j_1, \ldots, j_k$ are all distinct so they form a permutation of $1, \ldots, k$. – Daniel Schepler Sep 27 '17 at 22:43
• I see!! At the sum $\sum_{\pi\in \text{Per}_k}$ is the permutation of the form $\pi: (1, \ldots , k)\mapsto (j_1, \ldots , j_k)$ ? Or can we not say anything abou that? @DanielSchepler – Mary Star Sep 27 '17 at 22:46
• Yes, that's right. – Daniel Schepler Sep 27 '17 at 22:55