Determinant of a dual map.

Question

I've been trying to solve this problem but i don't know how to proceed. I found a proof online but i really don't understand it.

Problem: Prove that $\det(T) = \det(T^*)$, where $T: V \to V$ is a linear map and $T^*$ is the dual of the map $f$, i.e. $T^*: V^* \to V^*$ and $T^*(f) = f \circ T$. Suppose that$ V$ is finite dimensional.

Proof: Let be $e_1,\cdots,e_n$ base of $V$ and $w_1,\cdots,w_n$ base of $V^*$

(the proof is in the image)

Thank you so much.

Answer 1

for everyone that is looking for the same problem, i found the answer quite simple (without using tensors or wedge product). You have to proove that

$[T]\quad=\quad{[T*]}^{T}$

since

$det(A)\quad =\quad det({ A }^{ T })$

There's actually another question about how to find a matrix representation of a dual map called

Matrix Representation of a Dual Map

I hope it was helpfull.