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This will likely seem a bit contrived, and admittedly it is, but I wanted to see just how "close" we could get to generalizing the concept of a determinant. In what follows, we will lose quite a few desirable properties of the determinant as expected, since we cannot keep all of them. We will, however, manage to retain these properties for the case of square matrices.

Let $V, W$ be vector spaces over some field $\mathbb{F}$, with $\dim(V)=n$, $\dim(W)=m$. Let $T:V \rightarrow W$ be a linear transformation and let $\Lambda^p V$ be the $p$-th exterior power of $V$, with $\Lambda^p W$ defined analogously. Let $B= \left\{e_i \right\}_{i=1}^n$ be a basis for $V$ and let $B'= B- B \cap \ker(T)$. Choose an ordered basis $A \subseteq W$ such that $T(B')^* \subseteq A$, where $T(B')^*$ denotes the largest set of linearly independent elements of $T(B')$. Define $\alpha= \displaystyle \bigwedge_{e_i \in A- T(B')} e_i$, where the product respects the order on $A$ (If $A- T(B')= \emptyset$, we obey the empty product convention and let $\alpha=1$).

Define the following linear maps:

$\ell: \Lambda^n V \rightarrow \Lambda^n W$, $\ell: v_1 \wedge ... \wedge v_n \mapsto T(v_1) \wedge ... \wedge T(v_n)$

$g: \Lambda^n W \rightarrow \Lambda^m W$, $g:b \mapsto \alpha \wedge b$

And finally: $\gamma: \Lambda^n V \rightarrow \Lambda^m W$, $\gamma: a e_1 \wedge...\wedge e_n \mapsto (g \circ \ell)(ae_1 \wedge ... \wedge e_n)$

$\gamma$ is a composition of vector space homomorphisms (linear maps) and hence is also a vector space homomorphism. Now $\dim(\Lambda^n V)= \dim(\Lambda^m W)=1$, so there is some $c \in \mathbb{F}$ such that $\gamma: ae_1 \wedge ... \wedge e_n \mapsto ca e_1' \wedge ... \wedge e_m'$ (where $e_i' \in A$), since $\gamma$ is linear. $\textit{Define}$ $c$ as the determinant with respect to bases $A,B$ and write: $c = \det_{A,B}(T)$.

To show that this coincides with the usual determinant for square matrices, we need only check the case $W=V$, in accordance with the definition given here.

First suppose $T: V \rightarrow V$ is a surjection. Then $\dim($im$(T))=\dim(V)$, so $\dim(\ker(T))=0$ by rank-nullity, so $T$ is in fact a bijection. Then $\alpha=1$, hence $g$ is the identity on $\Lambda^nV$. Then $\gamma= \ell$ and our definition of the determinant is precisely the same as the usual one.

Now suppose $T$ is not a surjection. Then $\dim($im$(T))<n$, so there is some $e_i \in B$ with $T(e_i)= \displaystyle \sum_{j \neq i} c_j T(e_j)$ for $c_j \in \mathbb{F}$. Then $\ell(e_1 \wedge ... \wedge e_n)=0$, hence $\ell$ is trivial and thus $\gamma$ must be as well, i.e. we then have $c=0$. This also coincides with the usual definition of the determinant, since if $T$ is not surjective, it is not invertible, so we have $\det(T)=0=c$.

My question: Is this a valid generalization? We lose quite a bit here, we have to specify two bases and the order matters (it determines the sign), yet if the above is correct we still manage to retain the original definition for square matrices. A little work shows that if $m<n$, $\det(T)=0$, which isn't always the case for $m>n$, hence we immediately lose $\det_{A,B}(C)= \det_{A,B}(C^T)$.

My Second Question: If this generalization is valid, does anyone have any ideas as far as possibly making this coordinate independent? I really couldn't find a way.

My Third Question: Admittedly more subjective, but does this seem natural?

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This only answers question 2: There is no hope to do anything independent of a choice of bases here. If you have two linear maps $f,g:V\to W$, which have the same rank, then there are linear isomorphisms $S:V\to V$ and $T:W\to W$ such that $g=T\circ f\circ S$. Equivalently, you can choose bases such that the two maps correspond to the same matrix. To see this, suppose that the dimensions of $V$ and $W$ are $n$ and $m$ and that $f$ has rank $r$. Then choose a basis $v_1,\dots,v_n$ for $V$ such that $v_{r+1},\dots ,v_n$ form a basis for the kernel of $f$. Then $w_1:=f(v_1),\dots, w_r:=f(v_r)$ form a basis for the image of $f$, and you arbitrarily extend to a basis $\{w_1,\dots, w_m\}$ for $W$. Then $f(v_i)=\begin{cases} w_i\quad i\leq r\\ 0 \quad i>r\end{cases}$, which gives you a standard matrix form.

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  • $\begingroup$ Yeah I suspected there would be no way. $\endgroup$ – M10687 Jun 9 '16 at 13:29
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Here is what they actually do. It is a worthwhile exercise to confirm that everything is well-defined, that changing basis in the underlying vector space does not alter anything.

enter image description here

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