Given a finite $K$-vector space $E$ of dimension $n$ and a linear map $f:E\rightarrow E$, one can define the determinant of $f$ as the unique scalar $\lambda$ satisfying :
For any multilinear alternating form $\phi :E^n\rightarrow K$ and for all $x_1,...,x_n \in E$, $\phi (f(x_1),...,f(x_n))=\lambda \phi(x_1,...,x_n)$.
And we define the transpose $^t \!f$ of $f$ as follows :
$^t\!f:E^*\rightarrow E^*,\; ^t\!f(u)=u\circ\ f$.
My question is : is it possible, using this definition, to prove that $\det(f)=\det(^t\!f)$ without using matrices or the expression of $f$ in a base of $E$ ?
If we consider an inner product $(.|.)$ over $E$, I managed to prove that $\det(f)=\det(f^*)$ where $f^*$ is the adjoint operator of $f$ : if we define $\psi :E\rightarrow E^*$ by $\psi(x)(y)=(x|y)$ and $g:E^n\rightarrow K,\; (x_1,...,x_n)\mapsto \phi(\psi(x_1),...,\psi(x_n))$ ($\phi$ is a given $n$-linear alternating form over $E$) then g is an alternating multilinear form and $\forall v_1,...,v_n \in E$
\begin{align}
\det(^t\!f)g(v_1,...,v_n)&=\det(^t\!f)\phi(\psi(v_1),...,\psi(v_n))\\
&=\phi(^t\!f(\psi(v_1)),...,^t\!\!f(\psi(v_n)))\\
&=\phi(\psi(f^*(v_1)),...,\psi(f^*(v_n)))\\
&=g(f^*(v_1),...,f^*(v_n))\\
& =\det(f^*)g(v_1,...,v_n)\\
\Rightarrow \det(f)=\det(f^*)
\end{align}
So we just have to prove that $\det(f)\det(f^*)$. I don't whether this makes things easier or not,but I'd be grateful if you could help me out. Thanks in advance.
Show that $f\mapsto \det(\,{}^t\!f)$ is an alternating multilinear form and see what happens when $f=\operatorname{id}$.