Defining the transpose of a linear map "directly" in terms of the linear map I am wondering if there is anything wrong with the following way of defining the transpose of a linear map, at least for finite dimensional vector spaces. (The usual definition I'm aware of does it by looking at dual spaces/dual maps, and I'm wondering if sometimes we don't really need to consider that, and perhaps just "do it more directly" sometimes.)
Let $V, W$ have bases $\{ v_1, ..., v_n\}$ and $\{w_1, ..., w_m\}$, respectively. Let $M_1: \mathcal{L}(V,W) \to \mathbb{F}^{m,n}$ be the isomorphism that takes a linear transformation to its matrix, and similarly for $M_2: \mathcal{L}(W,V) \to \mathbb{F}^{n,m}$. Let $t$ denote the operation of taking the transpose of a matrix (which is an isomorphism $\mathbb{F}^{m,n}$ to $\mathbb{F}^{n,m}$), so for example $t(A)$ is the transpose of $A$, which we also denote by $A^t$.
Then suppose I define the transpose of a linear map $T \in \mathcal{L}(V,W)$ by $T' := M_2^{-1}(t(M_1(T))) = M_2^{-1}(M_1(T)^t)$. Basically, we take a linear map, find its matrix with respect to the given bases, transpose the matrix, then get the linear map corresponding to this new matrix.
We can explicitly find $T'$ if we know what $T$ does, i.e. if we know what the matrix $M_1(T)$ is: Let $T'$ be the unique linear map sending each $w_k$ to $(M_1(T)^t)_{1, k}v_1 + ... + (M_1(T)^t)_{n, k}v_n$. Then $M_2(T') = M_1(T)^t$ (or equivalently $T' = M_2^{-1}(M_1(T)^t)$). So indeed the matrix of $T'$ is the transpose of the matrix of $T$, which captures the idea we intended.
The apparent benefit of this approach seems to be that we've defined the transpose of a linear map directly in terms of the transpose of matrices, and we used the original bases instead of considering dual bases. Also we can find the map $T'$ explicitly if we know the matrix of $T$.
Is this approach useful at all? Did I make any conceptual/logical errors?
 A: Let $V$ and $W$ be one-dimensional spaces, each spanned by the vectors $v$ and $w$ respectively, and let $T \colon V \to W$ be the linear operator $Tv = w$.
Using the bases $(v)$ and $(w)$, the matrix of $T$ is $[1]$, so when we transpose it and re-interpret it as the matrix of a map $W \to V$, we get $T^t \colon W \to V$ is the map $T^t(w) = v$.
However, using the bases $(2v)$ and $(w)$, the matrix of $T$ is $[2]$, so transposing and re-interpreting it as a map $W \to V$ gets $T^t(w) = 2v$, which is a different linear operator to before.
This definition of a transpose depends on the bases chosen for $V$ and $W$. The reason that we normally define the transpose as a map $T^* \colon W^* \to V^*$ between dual spaces is that it doesn't depend on whatever bases are chosen: the definition is simply that $(T^* f)(v) = f(Tv)$ for any $f \in W^*$ and $v \in V$.
When you pick bases for $V$ and $W$ you define isomorphisms $\varphi \colon V \to V^*$ and $\psi \colon W \to W^*$ (taking the basis to its dual basis), and your construction is equivalent to defining $T^t = \varphi^{-1} \circ T^* \circ \psi$. So there is not a uniquely defined $T^t$ in the same sense that $T^*$ is uniquely defined, because $T^t$ depends on the bases chosen. By picking bases appropriately we can actually make $T^t \colon W \to V$ into absolutely any linear operator $W \to V$, provided that it has the same rank as $T$, so this is not a good notion of a transpose operator. (If $V = W$ so $\varphi = \psi$ then $T^t$ could be any operator which is similar to $T$, which is slightly better, but still unable to be used as a transpose).
