If we have an isomorphism between vector spaces $f$, we can define an isomorphism whom representative matrix is the inverse of the $f$'s one in every basis (this is clearly $f^{-1}$). I was wondering if we could do the same also for transpose. I know for example that given a linear application $f:V \to W$, we can define $f^t:W^* \to V^*$ and this linear application's representative matrix is the transpose of $f$'s one but only in dual basis .And I know that(in the real case) the adjoint's representative matrix is the transpose of the $f$'s one, but only in orthonormal basis.(That is practically the same of the last case, since a dual basis authomatically induces a scalar product $\langle v_i,v_j \rangle=v_i^ *(v_j)$). Why do we need always some extra structure? Isn't there a more natural way to define this?
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1There is not a more natural way to do this because the is no natural isomorphism between $V$ and its dual space $V^*$. Every identification of $f^t$ with a map from $W$ to $V$ requires a basis in some sense – Ben Grossmann Jul 16 '20 at 14:42
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1You may find this post interesting – Ben Grossmann Jul 16 '20 at 15:32