Proposition

Let $e_1, \cdots, e_n$ be the usual basis of $\mathbb{R}^n$ and let $\phi_1, \cdots, \phi_n$ be the dual basis.

  1. Show that $\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}(e_{i_1}, \cdots, e_{i_k}) = 1$. What would the right side be if the factor $(k + l)!/k!/l!$ did not appear in the definition of $\wedge$?
  2. Show that $\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}(v_1, \cdots, v_k)$ is the determinant of the $k \times k$ minor of \(\begin{bmatrix} v_1 \\ \vdots \\ v_k \end{bmatrix}\) by selecting columns $i_1, \cdots, i_k$.

Solution

1

First, we claim that $\phi_{i_1} \wedge \cdots \wedge \phi_{i_k} = k! \Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})$ for any $1 \leq k \leq n$. If $k = 1$, then this is obvious. Suppose we showed this for some $1 \leq k \leq n - 1$.

\[\begin{align*} (\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}) \wedge \phi_{i_{k + 1}} &= \frac{(k + 1)!}{k!1!}\Alt((\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}) \otimes \phi_{i_{k + 1}}) \\ &= (k + 1)\Alt((\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}) \otimes \phi_{i_{k + 1}}) \\ &= (k + 1)\Alt(k!\Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k}) \otimes \phi_{i_{k + 1}}) \\ &= (k + 1)!\Alt(\Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k}) \otimes \phi_{i_{k + 1}}) \\ &= (k + 1)!\Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k} \otimes \phi_{i_{k + 1}}). \end{align*}\]

Therefore, the claim is true for any $k$.

Let $k, i_1, \cdots, i_k$ be given.

\[\begin{align*} (\phi_{i_1} \wedge \cdots \wedge \phi_{i_k})(e_{i_1}, \cdots, e_{i_k}) &= k!\Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(e_{i_1}, \cdots, e_{i_k}) \\ &= \frac{k!}{k!}\sum_{\sigma \in S_k} \sgn(\sigma) (\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(e_{i_{\sigma_1}}, \cdots, e_{i_{\sigma_k}}) \\ &= (\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(e_{i_{1}}, \cdots, e_{i_{k}}) \\ &= 1. \end{align*}\]

Suppose that the factor $(k + 1)!/k!l!$ does not appear in the definition of $\wedge$. Then $\phi_{i_1} \wedge \cdots \wedge \phi_{i_k} = \Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})$. This can be shown by induction because

\[\begin{align*} (\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}) \wedge \phi_{i_{k + 1}} &= \Alt((\phi_{i_1} \wedge \cdots \wedge \phi_{i_k}) \otimes \phi_{i_{k + 1}}) \\ &= \Alt(\Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k}) \otimes \phi_{i_{k + 1}}) \\ &= \Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_{k + 1}}). \end{align*}\]

Let $k, i_1, \cdots, i_k$ be given.

\[\begin{align*} (\phi_{i_1} \wedge \cdots \wedge \phi_{i_k})(e_{i_1}, \cdots, e_{i_k}) &= \Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(e_{i_1}, \cdots, e_{i_k}) \\ &= \frac{1}{k!}\sum_{\sigma \in S_k} \sgn(\sigma) (\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(e_{i_{\sigma_1}}, \cdots, e_{i_{\sigma_k}}) \\ &= \frac{1}{k!}(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(e_{i_{1}}, \cdots, e_{i_{k}}) \\ &= \frac{1}{k!}. \end{align*}\]

2

Each $v_i$ can be written as $\sum_{j=1}^{n} a_{ij} e_j$.

\[\begin{align*} \phi_{i_1} \wedge \cdots \wedge \phi_{i_k} (v_1, \cdots, v_k) &= k! \Alt(\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(v_1, \cdots, v_k) \\ &= \sum_{\sigma \in S_k} (\phi_{i_1} \otimes \cdots \otimes \phi_{i_k})(v_{\sigma_1}, \cdots, v_{\sigma_k}) \\ &= \sum_{\sigma \in S_k} \phi_{i_1}(v_{\sigma_1}) \cdots \phi_{i_k}(v_{\sigma_k}) \\ &= \sum_{\sigma \in S_k} a_{i_1, \sigma_1} \cdots a_{i_k, \sigma_k}. \end{align*}\]

The last expression is exactly the determinant of the $k \times k$ minor of the matrix in the problem statement.