Pullback in the top dimension

Question

I am solving exercises from Loring Tu.

Show that if $L : V \rightarrow V$ is a linear operator on a vector space V of dimension n, then the pullback $L^{\wedge} : A_n(V) \rightarrow A_n(V)$ is multiplication by determinant of L.

Attempt:

This is a linear operator between the same spaces of the dimension 1. It follows that it must be multiplication by a constant. I don't understand why is it the determinant ?

Answer 1

Here are some options:

Pick a basis $v_1, \ldots, v_n$ for $V$.

Approach 1: The vector $d=v_1^*\wedge \ldots \wedge v_n^*$ is a basis for $A_n(V)$, and $L^{\wedge}(d)=L(v_1)^*\wedge \ldots \wedge L(v_n^*)$. Now if $L$ has matrix $M$ with entries $m_i^j$ in this basis, i.e. $L(v_i)=\sum_j m_i^j v_j$ then $L^{\wedge}(d)=(\sum_{j_1} m_1^{j_1} v_{j_1}) \wedge \ldots \wedge (\sum_{j_n} m_n v^{j_n})$, which you can rewrite as sum over permutations by property of wedge product, recovering precisely the sum over permutations formula for $det M$ - times $v_1^*\wedge \ldots \wedge v_n^*$. Thus the rescaling factor is $\det M$ (aka $\det L$).

Approach 2: Let $^{\wedge}$ be the map that sends any linear map $L:V\to V$ to the rescaling factor of $L^{\wedge}$. Since we have fixed a basis for $V$, we get a map sending matrices to numbers. Then we check that this map: 1) is linear in each column of $M$, 2) is totally antisymmetric in permuting the columns of $M$ and 3) sends the identity matrix to $1$. Then a) this determines such a map from matrices to numbers uniquely, and b) $M\to \det M$ has all these properties. Thus $^{\wedge}$ sends $L$ to the determinant of $M$ (aka the determinant of $L$).

Answer 2

Choose a basis $v_1,\dots,v_n$ for $V$, and consider the linear map $\phi : \mathbb R^n \to V$ given by $\phi(a_1,\dots,a_n) = a_1v_1 + \cdots + a_nv_n$. Now, for $\omega \in A_n(V)$ define $\eta_\omega \in A_n(\mathbb R^n)$ by $$\eta_\omega(x_1,\dots,x_n) = \omega(\phi(x_1),\dots,\phi(x_n)).$$ Since $\det \in A_n(\mathbb R^n)$ form a basis for $A_n(\mathbb R^n)$, there exists $c \in \mathbb R$ such that $\eta_\omega = c \det$. Evaluating in the standard basis for $\mathbb R^n$ we see that $c = \omega(v_1,\dots,v_n)$, and so we have \begin{align} L^\vee(\omega)(v_1,\dots,v_n) &= \omega(Lv_1,\dots,Lv_n) \\ &= \eta_\omega(\phi^{-1}(Lv_1),\dots,\phi^{-1}(Lv_n)) \\ &= \det(\phi^{-1}(Lv_1),\dots,\phi^{-1}(Lv_n)) \omega(v_1,\dots,v_n) \\ &= (\det L) \omega(v_1,\dots,v_n) \end{align} meaning that $L^\vee(\omega) = (\det L) \omega$.