The condition $\det(I)=1$, and the uniqueness of the determinant

Question

The third condition for the definition of a determinant is given as $\det(I)=1$ (multi-linear and alternating are the first two conditions). Why is this important? Does it relate to the uniqueness of the determinant? What exactly is meant by uniqueness?

Answer 1

It is necessary to have $\det(I)=1$ in order for $\det$ to be not totally trivial ($0$ for every matrix) and still satisfy the property $\det(AB)=\det(A)\det(B)$. Observe that if $\det(I)=a$ for some number $a\neq 0,1$, then $$\det(I^2)=\det(I)=a\neq a^2=\det(I)\det(I)$$ The property $\det(AB)=\det(A)\det(B)$ is extremely important, both theoretically (it makes the function $\det:\mathrm{GL}_n(K)\to K^\times$ a group homomorphism) and computationally.

With regards to your question about uniqueness: one common way of defining the determinant (which it sounds like is the way it's the way you're using) is to first propose considering functions $f:\mathrm{M}_{n\times n}(K)\to K$ that are (a) multilinear, (b) alternating, and (c) have $f(I)=1$; and then showing that in fact there is a unique such function, and we give that function the name "determinant".