Prove $\det(T^*)=\det(T)$ using exterior algebta

Question

Let $T:V\to V$ be a linear operator on a finite-dimensional vector space $V$. $T^*:V^*\to V^*$ is the transpose of $T$ (i.e., $T^*(\lambda)=\lambda\circ T$). I want to find a coordinate-free proof of $\det(T^*)=\det(T)$. Does anyone know a proof?

Answer 1

Hint This follows almost immediately from the coordinate-free characterization of $\det$: The map $T$ determines a canonical map $$\det T : \Lambda^n V^* \to \Lambda^n V^*$$ defined by extending the map $v_1 \wedge \cdots \wedge v_n \mapsto T(v_1) \wedge \cdots \wedge T(v_n)$ by linearity; here $n := \dim V$. (The existence of this map---and its well-definedness---are guaranteed by the universal property.) Since $\dim \Lambda^n V^* = 1$, $\det T$ is just multiplication by some constant in the base field of $V$, and this is precisely the determinant.

Additional hint More-or-less by construction $\det (T^*) = (\det T)^*$, but $\det T$ and $(\det T)^*$ are adjoint maps from a $1$-dimensional vector space to itself, so they are both multiplication maps by the same constant.

Answer 2

Suppose that $V$ is $n$-dimensional, the space of $n$-alternating forms defined on on $V$ has dimension 1 and $det$ is a generator. Let $f(A)=det(A^*)$ $A$ is multilinear and alternating. Thus there exists $c$ such that $det(A^*)=cdet(A)$. In particular, if $A=I_n$ the matrix of the identity map, $I^*=I$ thus $det(I^*)=cdet(I)$ implies $c=1$ and for every $A$, $det(A^*)=det(A)$.