# Definition of the determinant as of a function $d:\operatorname{End}(V)\rightarrow\mathbb{R}$

I'm trying to characterize the determinant in the following way. Consider a finite-dimensional vector space $$V$$ over $$\mathbb{R}$$ and the natural ring of its endomorphisms $$\operatorname{End}(V)$$. Given the determinant, we may define a function $$d:\operatorname{End}(V)\rightarrow\mathbb{R}$$ with the following properties:

1. $$d(1)=1$$
2. $$d(fg)=d(f)d(g)$$
3. $$d(f)\neq0\iff f$$ has a two-sided inverse (i.e. $$f$$ is an automorphism)

Can we complete this set of conditions in such a way that the determinant could be defined. Is there some category theoretical point of view?

• They don't. Take some power of the det instead. Jan 10, 2021 at 19:52
• Over $\mathbb{R}$ any positive real power of the absolute value of the determinant also satisfies these conditions. If you add the condition that $d$ is a polynomial then you get exactly the positive integer powers of the determinant. There are many different possible characterizations from here. Jan 10, 2021 at 21:06

If you add the condition that $$d$$ is an alternating multilinear map of $$f(e_1), \dots, f(e_n)$$ where $$\{e_1, \dots, e_n\}$$ is the canonical basis of $$V$$ then you’ll get that $$d$$ is the determinant.
• Worth noting though that this condition together with $d(1)=1$ is already sufficient to characterize the determinant, so conditions 2. and 3. from the question are not necessary anymore. Jan 10, 2021 at 20:34