Transpose of a linear operator is well-defined I would like to define the transpose of a linear operator $T$ between finite-dimensional vector spaces $V$ and $W$. Wikipedia gives a straightforward one, but I would like to define it based on what I already know about matrices. Thus, I define the transpose of $T$ to be given by choosing bases $B_1$ and $B_2$ of $W$ and $V$ respectively, and taking the transpose of $T$'s matrix with respect to these bases.
I would like to show that this definition is independent of basis choice, however. In other words, if the matrix chosen above is $A$, then if I changed basis from $B_1$ with change of basis matrix $Q$ and from $B_2$ with matrix $P$, that $QA^TP^{-1} = (PAQ^{-1})^T$. This does not seem to be the case, however. What have I done incorrectly, if anything?
 A: As mentioned in the comment, such a transpose map on $W$ and $V$ is coordinate dependent.
If you want a coordinate independent map, you'll have to stick with the map from $W^*$ to $V^*$. 
For a finite dimensional vector space, the double dual space of $V$ (i.e. $V^{**}$) is naturally isomorphic to $V$ (natural here has a technical definition, but essentially means this isomorphism does not depend on coordinates). 
On the other hand, even though the dual space of $V$ (i.e. $V^*$) and $V$ are isomorphic, we do not have that they are naturally isomorphic. Picking an isomorphism is nearly the same thing as selecting an inner product (i.e. adding geometry to $V$). 
Given a linear map, $A:V \to W$, we can define $A^T:W^* \to V^*$ naturally via $A^T(f)(v)=f(A(v))$. [$f:W \to \mathbb{R}$ is a linear functional on $W$, $A^T(f):V \to \mathbb{R}$ is a linear functional on $V$.]
If we want to define $A^T$ as a map from $W$ to $V$, we'll need to pass (somehow) from $W^*$ to $W$ and $V^*$ to $V$. One way to do this is via inner products. 
Suppose that $\langle v_1,v_2 \rangle_V$ is an inner product on $V$ and $\langle w_1,w_2 \rangle_W$ is an inner product on $W$. Then $v \mapsto \langle v, \cdot \rangle_V$ gives an isomorphism from $V$ to $V^*$. Let's give this a name: $\varphi_V(v)=\langle v, \cdot \rangle_V$. Likewise, $w \mapsto \langle w,\cdot \rangle_W$ from $W$ to $W^*$ denoted $\varphi_W(w)=\langle w,\cdot \rangle_W$.
Then we have: $A^T(\varphi_W(w))(v) = (\varphi_W(w))(A(v)) = \langle w, A(v) \rangle_W$. If we then use $\varphi_V^{-1}$ we can turn the linear map $A^T(\varphi_W(w))$ into a vector in $V$. Thus composing maps as follows: $\varphi_V^{-1} \circ A^T \circ \varphi_W$ gives us your desired transpose map:
$$\varphi_V^{-1} \circ A^T \circ \varphi_W:W \to V$$
If we call this $\widetilde{A^T} = \varphi_V^{-1} \circ A^T \circ \varphi_W$, we have
$$ \langle \widetilde{A^T}(w),v \rangle_V = \langle w, A(v) \rangle_W$$
for all $v \in V$ and $w \in W$. Some texts would just use the above equality as the definition of the transpose map (but, of course, this depends on those pesky inner products).
The issue is that turning the transpose map into a map on $W$ and $V$ requires us to pick isomorphisms between $V$ and its dual as well as $W$ and its dual. A special case of this is selecting an inner product. 
Picking isomorphisms between vector spaces and their duals is very closely related to selecting bases [Given a basis for $V$, you get a dual basis for $V^*$ and bam you have an isomorphism.]
Likewise, picking an inner product is closely related to selecting a basis [Pick a basis,  declare it to be orthonormal, and bam you have an inner product.]
In the end, you really can't get the transpose map to be "coordinate free" on the original vector spaces in the way you're looking for. It all comes down to the fact that a vector space and its dual are not naturally isomorphic.
