Proposition

If $f: V \rightarrow V$ is a linear transformation and $\dim V = n$, then $f^{*}: \Lambda^n(V) \rightarrow \Lambda^n(V)$ must be multiplication by some constant $c$. Show that $c = \det f$.

Solution

$\Lambda^n(V)$ is a 1-dimensional vector space. Since $f^*$ is a linear map, it must be multiplication by some constant $c$.

Let $e_1, \cdots, e_n$ be a basis of $V$ and $\phi_1, \cdots, \phi_n$ be the dual basis.

As we showed here, $(\phi_1 \wedge \cdots \wedge \phi_n)(e_1, \cdots, e_n) = 1$. Therefore, $c = c(\phi_1 \wedge \cdots \wedge \phi_n)(e_1, \cdots, e_n)$. Then we have $c = f^*(\phi_1 \wedge \cdots \wedge \phi_n)(e_1, \cdots, e_n) = (\phi_1 \wedge \cdots \wedge \phi_n)(f(e_1), \cdots, f(e_n))$. By Theorem 4.6 [Spivak], this equals $\det(f) \cdot (\phi_1 \wedge \cdots \wedge \phi_n)(e_1, \cdots, e_n) = \det(f)$.

Therefore, $c = \det(f)$.