Proposition 10

Let $M$ and $N$ be two free $A$-modules of respective dimensions $m$ and $n$, $u : M \to N$ a linear mapping and $X$ the matrix of $u$ with respect to a basis $(e_i)_{1 \leq i \leq m}$ of $M$ and a basis $(f_j)_{1 \leq j \leq n}$ of $N$. Then, for every integer $p \leq \inf(m, n)$, the matrix of $\wedge^p(u)$ with respect to the basis $(e_K)_{K \in \mathfrak{F}_p(I)}$ of $\wedge^p(M)$ and the basis $(f_H)_{H \in \mathfrak{F}_p(J)}$ of $\wedge^p(N)$ (where we have written $I = \{1, m\}$ and $J = \{1, n\}$) is the matrix $(\det(X_{H,K}))$ of type $(\mathfrak{F}_p(J), \mathfrak{F}_p(I))$ (and hence with $\binom{n}{p}$ rows and $\binom{m}{p}$ columns).

For a subset $K \subset J$ with $p$ elements, let $(j_k)_{1 \leq k \leq p}$ be the sequence of elements of $K$ arranged in increasing order; by definition of $\wedge^p(u)$, by § 7, no. 2, formula (4)

Hence the element of the matrix of $\wedge^p(u)$ which is in the row of index $H$ and the column of index $K$ is the component of index $H$ of the element $u(e_{j_1}) \wedge \cdots \wedge u(e_{j_p})$; it is therefore equal to $\det(X_{H,K})$ by Proposition 9.

The matrix $(\det(X_{H,K}))$ is called the $p$-th exterior power of the matrix $X$ and is denoted by $\wedge^p(X)$. When $p = m = n$, $\wedge^n(X)$ is the matrix with the single element $\det(X)$.