I was reviewing some school notes from many semesters ago and I came across a point which I wish to prove but can't.

Let $F$ be a field (real or complex for example), and we say $\delta : Mat(n,n,F)\to F$ is an alternating multilinear map along the rows if: (1) it is a multilinear map with the n rows being the n coordinates of the map (2) switching any two rows reverses the sign of the result.

My notes then wrote "know the theorem which says that if $\delta (I)=1$, then $\delta(AB)=\delta(A)\delta(B)$". I wish to show that this is true, since now I am not happy with just "knowing" this fact.

[The prof then showed us that $det$ is the unique alternating multilinear map along the rows such that $det(I)=1$]


Okay, it appears that I figured out a way around this after lunch. I can prove that determinant is the only alternating multilinear map such that $det(I)=1$ without using the theorem that I want to prove, and then use the uniqueness of determinant to prove the theorem. Namely:

See theorem 3 of here to get uniqueness of determinant. Then basically we just need to prove $det(AB)=det(A)det(B)$. If $B$ is non-invertible, then so is $AB$, so the case $|B|=0$ is okay. Otherwise, define $\delta_B(A):=det(AB)/det(B)$ and show that $\delta_B$ is an alternating multilinear map with $\delta_B(I)=1$, so $\delta_B = det$.


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