Determinant of linear transformation

Question

Given a linear transformation $T:V\rightarrow V$ on a finite-dimensional vector space $V$, we define its determinant as $\det([T]_{\mathcal{B}})$, where $[T]_{\mathcal{B}}$ is the (square) matrix representing $T$ with respect to a basis $\mathcal{B}$. It is proven that this does not depend on the particular choice of the basis $\mathcal{B}$.

My question is:

Is there a similar definition of determinant for a linear transformation $T:V\rightarrow W$, where $V,W$ are finite-dimensional vector spaces with the same dimension?

Answer 1

You can define it either

a. with respect to two fixed bases $B_1$ of $V$ and $B_2$ of $W$ or

b. with respect to an isomorphism $\varphi : V\to W$.

In the latter case, if $B=\{v_1,\ldots,v_n\}$ is a basis of $V$, then
$\varphi(B)=\{\varphi v_1,\ldots,\varphi v_n\}$ is a basis of $W$, and the determinant is independent of the choice of $B$, provided that $Tu$ is analyzed in terms of $\varphi(B)$.

Answer 2

I originally wrote this as a comment, but now I think it should maybe be an answer, so here goes.

I would argue that no (reasonable) such definition is possible. Admittedly, this is a bold claim, and maybe somebody could produce a definition I would be happy with. But my reason for the claim is that if you do the "natural" thing, i.e. write down a matrix for $T$ with respect to a basis $\mathcal{B}_1$ of $V$ and a basis $\mathcal{B}_2$ of $W$ and then take its determinant, then the answer depends on these choices. Thus what you have defined is not a property of the map $T$.

If you fix an isomorphism $\varphi\colon V\to W$, then you could take the determinant of $(T,\varphi)$ by picking a basis $\mathcal{B}$ for $V$ and taking the determinant of the matrix of $T$ with respect to $\mathcal{B}$ and $\varphi(\mathcal{B})$ as Yiorgos suggests - this doesn't depend on $\mathcal{B}$ for the same reason as in the $V\to V$ case, but it does depend on $\varphi$. In fact, this is essentially what you do in the $V=W$ case, but there there is a canonical choice of $\varphi$, namely the identity map on $V$. For two non-equal vector spaces of the same dimension, there is no such preferred isomorphism.

Answer 3

Yes there is , I think you should have studied this first.

Let $T: V \to W$ and let $\mathcal{B_2}$ and $\mathcal{B_1}$ be the basis of $V,W$ resp. Notation for that is $det(T)=[T]^{{\mathcal{B_1}}}_{\mathcal{B_2}}$. Simply write basis images of elements of $\mathcal{B_1}$ in terms of $\mathcal{B_2}$, and then make the matrix of coordinates, as you do for $T:V\to W$.

Reference for more details is Linear algebra by friedberg, insel and spence section $2.2$. Here is one important Image...