Determinant of a linear transform between two different vector spaces with the same dimension

(Rewritten - the original post went off into unnecessary territory, but I had leave and didn't have time to prune it appropriately then.)

Determinants are only defined on transformations of a finite-dimensional vector space into itself. They are not defined for transformations into another vector space, not even if that vector space is isomorphic to the first.

Most commonly, determinants are first defined on square matrices. To define them on transformations from an $n$-dimensional vector space $V$ over the field $Bbb F$ (if you don't understand that, just consider $Bbb F$ to be either $Bbb R$ or $Bbb C$, depending on whether your vector space is real or complex), you first find a basis for $V$. By that basis, you can identify every vector $v in V$ with an ordered $n$-tuple $(v_1, v_2, ..., v_n) in Bbb F^n$. This identification defines an invertible linear transformation $R : V to Bbb F^n$. If $T : V to V$ is a linear transformation, then $RTR^{-1}=Rcirc T circ R^{-1}$ defines a linear transformation from $Bbb F^n to Bbb F^n$. I.e., $RTR^{-1}$ is an $n times n$ matrix. As such its determinant is already defined. So we define $det T := det(RTR^{-1})$.

Now extrinsically, that definition depends on $R$ - which itself depends on the basis originally chosen. But it can be shown that the exact same value for $det T$ is obtained for any basis, so in fact $det T$ does not depend on the basis.

But one must be careful here: we make use of $R$ twice. Once to convert from $Bbb F^n$ to $V$ and then convert back from $V$ to $Bbb F^n$. Suppose we decided to choose different bases for those two conversions. We could try $RTS^{-1}$ where $S$ is defined by some other basis. But while it can be shown that $det STS^{-1} = det RTR^{-1}$, in general $det RTS^{-1} ne det RTR^{-1}$. For example, if the vectors for the basis defining $R$ happen to be $2$ times the vectors in the basis defining $S$, then $det RTS^{-1} = 2^ndet RTR^{-1}$. So the magic that allows you to get the same value of $det T$ regardless of the basis chosen requires the bases used for both transformations to be the same.

However, if $T : V to W$ with $V ne W$, then even when $V$ and $W$ are of the same dimension and therefore both isomorphic to $Bbb F^n$, you cannot choose the same bases for transforming $Bbb F^n to V$ and for transforming $W to Bbb F^n$. If $V$ and $W$ had the same basis, they would necessarily be the same vector space. Because we have no way of relating the two transformations $S : V to Bbb F^n$ and $R : W to Bbb F^n$, we cannot define a single $det T$ in this case. If $S'$ and $R'$ are tranformations defined by different bases, then in general, $det RTS^{-1} ne det R'TS'^{-1}$

So you can only talk about the determinant of a transformation $T : V to V$. The determinant of a transformation $T : V to W$ is not well-defined, even when $V$ and $W$ have the same dimension.