I'm interested in a coordinate-free proof of the statement $\mathrm{det}(A) = \mathrm{det}(A^T).$
Let $V$ be a finite-dimensional vector space over a field $K$, and let $f : V \rightarrow V$ be an endomorphism.
I define $\mathrm{det}(f)$ by $\mathrm{det}(f) \cdot \varphi = \varphi \circ f$, where $\varphi : V \times ... \times V \rightarrow K$ is any alternating function and $(\varphi \circ f)(v_1,...,v_n) := \varphi(f(v_1),...,f(v_n)).$
I define the transpose by $$f^* : V^* \rightarrow V^*, \; \psi \mapsto \psi \circ f.$$
I would like to prove that $\mathrm{det}(f) = \mathrm{det}(f^*)$ using these definitions but haven't been able to find a proof that doesn't involve choosing a basis. Thanks
For any simple $k$-vector $v_{(k)}$ and any simple $k$-form $\omega_{(k)}$, the linear operator $f$ and its adjoint $f^*$ should obey
$$\omega_{(k)}[f(v_{(k)})] = [f^*(\omega_{(k)})](v_{(k)})$$
where I have defined the action of a linear operator on $k$-vectors/$k$-forms as follows: if $v_{(k)} = v_1 \wedge v_2 \wedge \ldots \wedge v_k$ for $k$ linearly independent vectors, then
$$f(v_{(k)}) = f(v_1) \wedge f(v_2) \wedge \ldots \wedge f(v_k)$$
and similarly for the adjoint acting on a $k$-form.
It's not obvious to me that the above statements are immediately clear from your definition of the adjoint. It may be you have to prove it starting with the vector/1-form case:
$$\omega[f(v)] = [f^*(\omega)](v)$$
and go from there inductively. Once you have proved this notion, though, then the rest of the proof follows immediately, as $f(v_{(n)}) = (\det f) v_{(n)}$ and $f^*(\omega_{(n)}) = (\det f^*) \omega_{(n)}$.