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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?

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  • $\begingroup$ As some people stress in this discussion, I would like a definition intrinsic to the map $T$. $\endgroup$ – user181562 Oct 7 '14 at 16:08
  • $\begingroup$ I am not sure whether the question is still of interest. Be that as it may you can find precisely such a definition in the book Finite Dimensional Vector Spaces by Halmos. $\endgroup$ – Shahab Nov 11 '16 at 4:56
  • $\begingroup$ That is for the case $V=W$. $\endgroup$ – Shahab Nov 11 '16 at 5:04
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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 $\varphi(B)$.

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  • $\begingroup$ I think it's reasonable to argue, however, that you shouldn't do this, precisely because the determinant defined in this way is not a property of the map, but depends on a choice of basis for one of the two spaces (or an isomorphism of them). This isn't meant as a criticism of your answer, but I think it's worth pointing out - indeed, I would go so far as to say it means the answer to "is there a similar definition...?" is no! $\endgroup$ – mdp Oct 7 '14 at 15:50
  • $\begingroup$ I think Yiorgos' answer covers all bases. It might be worth pointing out that the answer in a. is basis dependent (that is, the value of the determinant depends on the bases chosen) and that b. addresses this shortcoming by 'tying' the $V,W$ bases together with $\phi$. $\endgroup$ – copper.hat Oct 7 '14 at 16:00
  • $\begingroup$ @copper.hat Yes, all the information is there (this is why I was trying not to sound critical, and I hope I succeeded!). I just thought it was worth drawing more attention to the dependencies. Part b. doesn't really address the shortcoming, it just shifts the dependency to $\varphi$ (although I think this more accurately explains what the necessary dependency is!). $\endgroup$ – mdp Oct 7 '14 at 16:04
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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.

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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...enter image description here

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  • $\begingroup$ The same dimension is needed so that you can take the determinant of the resulting matrix. Something is wrong where you write $\operatorname{det}(T)=[T]^{\mathcal{B}_1}_{\mathcal{B}_2}$; possibly there should be a $\operatorname{\det}$ on the right hand-side, but then this is undefined unless $\dim{U}=\dim{V}$. $\endgroup$ – mdp Oct 7 '14 at 15:47

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