Uniqueness of determinant function

Question

Prove that there can exist one only function $f: F^n \to F$ that satisfies the properties of multilinearity, antisymmetry, and normalization.

I cannot figure out how to get started. I would appreciate any help.

Answer 1

Here is a sketch. We basically want to show that given any matrix $A$, for any two functions $f, g$ satisfying those three axioms, $f(A) = g(A)$. We know that for any matrix $A$, there exist a sequence of invertible row operations $O_1, O_2, \cdots, O_k$ that when applied to $A$, bring it into reduced row-echelon form.

Now any reduced row-echelon form can be obtained via a sequence of operations $P_1, P_2, \cdots, P_l$ from the identity matrix. However, this time these operations may be elementary row operations or multiplying by zero. If $\text{rref}(A)$ is $n \times n$ and has $m \leq n$ pivot rows, then we can simply apply operations to $I_n$ until the first $m$ rows are equal to the first $n$ rows of $\text{rref}(A)$, and then multiply the remaining $n - m$ rows by zero.

It then follows that applying the sequence of operations $$P_1, P_2, \cdots, P_l, O_k^{-1}, O_{k-1}^{-1}, \cdots, O_1^{-1}$$ in that order, from left to right, will result in a sequence of operations transforming $I$ to $A$. However, each of these operations do one of three things:

1. Multiply a row by a constant $c$ (which may be zero)

2. Replace a row with itself plus another row

3. Switch two rows

all of which have a predictable outcome on the determinant. The first operation we know from multilinearity will multiply the determinant by $c$. The second operation we know won't change the determinant. Finally, from antisymmetry we know the third operation will negate the determinant. As $f(I) = g(I) = 1$, it must follow that $f(A) = g(A)$. $\square$