Determinant: Alternative Definitions Reference
Foundation for: Determinant: Continuity
Problem
Given a vector space $V$.
Consider an endomorphism $T:V\to V$.
The rank of an endomorphism:
$$\mathrm{rank}T:=\dim\left(\mathrm{im}T\right)$$
The determinant of an endomorphism:
$$\det T:=\text{???}$$
What would be a nice definition not relying on representations by matrices?
(I assume basic knowledge of Differential Geometry and Functional Analysis.)
 A: The nicest definition I know without getting into the symmetric group and matrix representation is the following:
Let $\;\lambda_1,...,\lambda_r\;$ be all the eigenvalues of $\;T\;$ (probably with repetitions and probably in some extension of the original definition field), then
$$\det T=\prod_{i=1}^r\lambda_i$$
A: This is an attempted simplification of the answer by Michael. The determinant of degree (?) $n$ is the unique multilinear alternating form $f:V^n\to k$ such that $f(\rm id)=1$. Here multilinear alternating means $f$ is linear in each coordinate and if we let $S_n$ act on $V^n$ by permuting the order of the vectors, $f((ij)x)=-f(x)$ for any $x\in V^n$. By $\rm id$ I mean $(e_1\mid e_2\mid \ldots \mid e_n)$. In particular, there exists a unique multilinear alternating form $f:V^n\to k$ such that $f(\rm id)=\alpha$ given $\alpha\in k$, which is $\alpha\det$. One can prove the uniqueness and existence by simple induction, although some cumbersome calculations are in order. 
Note that given any endomorphism $T:V\to V$, the mapping $T^*:V^n\to k$ defined by $T^*(v_1,\ldots,v_n)=\det(Tv_1,\ldots,Tv_n)$ is also multilinear alternating. Thus, there is a unique $\alpha$ such that $T^*=\alpha\det$, and we set $\alpha=\det T$.
A: Let $V$ be an $n$-dimensional vector space over the field $\mathbb{F}$. 
Given a linear map $T : V \to V$, there is an induced linear map $\bigwedge^nT : \bigwedge^n V \to \bigwedge^n V$ given by $\left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n)$. As $\bigwedge^nV$ is one-dimensional, $\bigwedge^nT = k\operatorname{id}_{\bigwedge^nV}$ for some scalar $k \in \mathbb{F}$. This scalar is precisely $\det T$.

Let me summarise some facts about the vector spaces $\bigwedge^pV$ (see the Wikipedia article on exterior algebras for more information).
Given a vector space $V$ of dimension $n$, there is an associated vector space $\bigwedge^pV$ for any $0 \leq p \leq n$ called the $p^{\text{th}}$ exterior power of $V$. The elements of $\bigwedge^pV$ are linear combinations of terms of the form $v_1\wedge\dots\wedge v_p$ where $v_1, \dots, v_p \in V$. The symbol $\wedge$ is called the wedge product, and it satisfies skew-symmetry, i.e. $v_i\wedge v_j = -v_j\wedge v_i$. If $\{v_1, \dots, v_n\}$ is a basis for $V$, then $\{v_{i_1}\wedge\dots\wedge v_{i_p} \mid i_1 < \dots < i_p\}$ is a basis for $\bigwedge^pV$ and therefore the dimension of $\bigwedge^pV$ is ${n \choose p}$.
A: There is no problem to write a definition of the determinant functional τ using induction by n and avoiding any mention of either matrices or n-forms. But it will not be very convenient since proof of uniqueness should rely on (concealed) decomposition of a matrix to upper and lower triangular factors.
This set of 3 axioms is sufficient to define unique τ on finite-dimensional spaces:
(Multiplicativity) $τ(AB) = τ(A)\,τ(B)$ for any endomorphisms A, B of V.
(Normalization) If $\dim V = 1$, then $τ(λI) = λ$.
(Decomposition) Let $U\subset V$ be a non-trivial (proper and ≠ {0}) A-invariant subspace (i.e. $\operatorname{im}(A\vert_U)\subseteq U$); we can then define the quotient space $W = V / U$ and such unique endomorphism $\hat A: W\to W$ that the diagram
 V → W
 ↓A   ↓$\hat A$
 V → W
commutes. Then $τ(A) = τ_U(A\vert_U)\,τ_W(\hat A)$ must hold, where right-hand side terms refer to τ defined on the subspace and the quotient space.
Of course, one might prove existence of τ for V of any finite dimension then, that is not a very easy task.
A: Let $V$ be a finite-dimensional linear space over a field $\mathbb{F}$. Let $\mathcal{L}(V)$ denote the linear operators from $V$ into itself. Then I would expect that the determinant to be the unique multiplicative function $\tau : \mathcal{L}(V)\rightarrow\mathbb{F}$ for which $\tau(I)=1$. That is, $\tau(AB)=\tau(A)\tau(B)$. Is it?
