Show that $c=\det f$ with $f: V \to V$ is a linear transformation.

Question

(Spivak 4-2.p.g 84). if $f: V \to V$ is a linear transformation and $\dim V=n$, then $f^\ast : \bigwedge^n(V) \to \bigwedge^n(V) $ must be multiplication by some constant $c.$ Show that that $c=\det f$.

My attempt, as $\dim \bigwedge^n(V)= 1$, then it is generated, just by a tensor, we choose $\omega$ (this ensures that for any $g \in \bigwedge^n(V)$, $g= c\omega$ with $c \in \mathbb R$ ).
As $f^\ast(g)$ it is an $n$-tensor in $V$, then it will be multiplication by some constant $c$ of $\omega$ as $\text{“}f^\ast(g)(v_1,...,v_n)=g(f(v_1),..,f(v_n))=c\omega(f(v_1),..,f(v_n))\text{''}$ with $v_1,...,v_n \in V$, can I think like this?, of anyone can give me some advice I appreciate it. “

Answer 1

Choose a basis $v_1,\dots,v_n$ for $V$. Then, for each $j \in \{1,\dots,n\}$ there exists $n$ real numbers $a_{1j},a_{2j},\dots,a_{nj}$ so that $f(v_j) = a_{1j}v_1 + a_{2j}v_2 + \cdots + a_{nj}v_n$, and then, by definition, the number $\det f$ is the determinant of the matrix $$\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{pmatrix}.$$ So, if $\omega \in \Lambda^n(V)$, then by theorem 4-6 (p.82) we have $$\omega(f(v_1),\dots,f(v_n)) = (\det f) \omega(v_1,\dots,v_n),$$ that is, $f^*(\omega)(v_1,\dots,v_n) = (\det f) \omega(v_1,\dots,v_n)$, meaning that $f^*(\omega) = (\det f)\omega$.