Consider $(d<\infty)$-dimensional ${ \Bbb R}$-vector spaces $V$ and $W$, and their dual spaces $V^*:=Hom(V,{\Bbb R})$ and $W^*:=Hom(W,{\Bbb R})$. Naturally, $V,W,V^*,W^*$ are all isomorphic. Now, the definition of a (linear) bijection $A:V\to W$ induces canonically a pullback operator $B:W^*\to V^*$ because $V^*$ and $W^*$ are functions on $V$ and $W$.
So far so good. My question is about from here on.
My questions:
The question is about the relation $B:=~^tA$ we enforce via the standard definition:
$$
(B(\alpha))(v):=\alpha(A(v)), \qquad v\in V, \alpha\in W^*.
$$
- Why we couldn't define instead: $$ (B(\alpha))(v):=(K(\alpha))(A(v)), \qquad K\in Aut(W^*), $$ resulting in a more general relation $B:=~^tA\circ K$?
And by such a definition,
- what are the consequences?
When we decide which vector in $V$ is which vector in $W$ ( i.e., $A$ is not canonic, but a choice), it seems rational that we also could decide (simultaneously) which function in $W^*$ is which function in $V^*$ (i.e., $B$ be a choice); of course by introducing a new isomorphism $K$.
A discussion:
So far I can see, such a definition alters the conventional invariance of scalar products $\alpha_{W^*}(w)=\alpha_{V^*}(v)$, to another relation $\alpha_{W^*}(w)=k_{K}\cdot\alpha_{V^*}(v)$ where $k_{K}\in {\Bbb R}$ depends on $K$. But is this make any trouble? Are there other changes?