This might be a dumb question; I know only enough group theory to be able to ask dumb questions.
Ken W. Smith has pointed out that one way to get intuition about the determinant is to observe that it maps matrix multiplication to real multiplication. As it is continuous, too, this means that it is a Lie group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to the multiplicative group of the nonzero reals, $\mathbb R^\times$. The natural question to ask, then, is whether it is the only such homomorphism.
Obviously not: any function $\mathbf A \mapsto (\det\mathbf A)^k$ for $k\in\mathbb Z$ is also a homomorphism.
But are all homomorphisms between the two groups of such a form? That is,
Is every Lie group homomorphism from $\mathrm{GL}_n(\mathbb R)$ to $\mathbb R^\times$ identical to $\mathbf A \mapsto f(\det\mathbf A)$ for some homomorphism $f:\mathbb R^\times\to\mathbb R^\times$?
Best Answer
In general, if $\phi : G \to H$ is a Lie group homomorphism then we get a Lie algebra homomorphism $\phi_* :\mathfrak g \to \mathfrak h$. If $\phi,\psi : G\to H$ are two Lie group homomorphisms then $\phi_* = \psi_*$ implies $\phi = \psi$. So we really just have to classify Lie algebra homomorphisms $\mathfrak{gl}_n(\mathbb R)\to \mathbb R$ and then see which ones lift to the groups.
Since the Lie algebra $\mathbb R$ is abelian, these Lie algebra homomorphisms are those elements of $\mathfrak{gl}_n(\mathbb R)^*$ that vanish on commutators. It turns out that this property uniquely characterizes the trace up to scale (I wasn't able to find a good reference for this but it should be easy, if a bit messy, to prove by hand).
So given $\phi : \mathrm{GL}_n(\mathbb R) \to \mathbb R^\times$, we have $\phi_* = c \operatorname{tr}$ for some $c \in \mathbb R$. Now $\phi_*$ restricts to the connected components where it integrates to $\mathrm{GL}_n(\mathbb R)_+ \ni A\mapsto(\det A)^c \in \mathbb R^+$. Of course for this homomorphism to extend to all of $\mathrm{GL}_n(\mathbb R)$ requires that the map $a \mapsto a^c$ be well-defined on all of $\mathbb R^\times$ so Lie group homomorphism is $\det$ followed by an automorphism of $\mathbb R^\times$.