Hello everyone I have the following statement :
So we have $E$ a vector space with $e_1 , ... , e_n$ as a basis , $\sigma$ a permutation of $ 1,2,...,n$. We are given an endomorphism on $E$ s.t $e_i \rightarrow e_{\sigma(i)}$.
Statement:
It's transpose is the endomorphism on $E^*$ (dual space) s.t :
$$e_i^* \rightarrow e_{\sigma^{-1}(i)}$$
I have no clue for how can I prove that it's true (I don't see why it's true) any hint would be appreciated.
Attempts:
I think that how the associated transpose is defined in $E^*$ may help , I've chosen $e^*_1,..,e^*_n$ as a basis for $E^*$ but it's going nowhere
Let $u_\sigma$ be the permutation endomorphism.
By definition of the transpose, we should have for all $f \in E^*$, $x \in E$
$$\langle ^tu_\sigma(f),x\rangle = \langle f,u_\sigma(x) \rangle.$$
In particular, we should have for all $i,j \in \{1, \dots, n\}$
$$\langle ^tu_\sigma(e_j^*),e_i\rangle = \langle e_j^*,u_\sigma(e_i) \rangle.$$
Let's see what we get if $^tu_\sigma=u_{\sigma^{-1}}$:
$$\langle ^tu_\sigma(e_j^*),e_i\rangle =\langle e_{\sigma^{-1}(j)}^*,e_i \rangle=\langle e_j^*,e_{\sigma(i)} \rangle= \langle e_j^*,u_\sigma(e_i) \rangle.$$
As the transpose is unique, we get the desired conclusion.