To demonstrate the power of differential forms, we consider some elementary applications. A map \begin{equation} A:V \longmapsto V \end{equation} is linear if it satisfies 1) \begin{equation} A(f\alpha + \beta) = f(A\alpha) + A\beta \end{equation} A linear map on 1-forms can be extended to a linear map on $p$-forms, also called $A$ (or occasionally $\bigwedge^p A$), given by \begin{equation} \alpha^1 \wedge … \wedge \alpha^p \longmapsto A\alpha^1 \wedge … \wedge A\alpha^p \end{equation} In particular, $A$ induces a linear map on $\bigwedge^n(V)$, which is 1-dimensional. But the only such map is multiplication by a scalar, which we call $|A|$. Thus, for any $\omega\in\bigwedge^n(V)$, we have \begin{equation} A(\omega) = |A|\,\omega \end{equation}

The linear map $A$ can be expressed in terms of a basis ${dx^i}$ as \begin{equation} A(dx^i) =: a^i{}_j dx^j \end{equation} where of course $\left(a^i{}_j\right)$ is precisely the matrix representation of $A$ with respect to the given basis. Working out the action of $A$ on the basis element $dx^1 \wedge … \wedge dx^n$ of $\bigwedge^n(V)$, it is easily seen that $|A|$ consists of a sum of all possible products of one element from each row and column of the matrix $\left(a^i{}_j\right)$, with suitable signs; $|A|$ is the determinant of $A$. Note however that the above definition is independent of basis!

We can now give an elegant derivation of the product rule for the determinant. Suppose $A$ and $B$ are linear transformations on $V$, and $\omega$ is any element of $\bigwedge^n(V)$. Then \begin{align} |A\circ B|\,\omega &= (A\circ B)\,\omega = A(B\,\omega) = A(|B|\,\omega) \nonumber\\ &= |B|(A\,\omega) = |B||A|\,\omega = |A||B|\,\omega \end{align} so that the determinant of a product is the product of the determinants.