Recall as in lecture our axioms for the volume of a parallel

Recall as in lecture our axioms for the volume of a parallelepiped: Given upsilon_1,...,upsilon_n vectors in R^n we have vol(upsilon, ...,upsilon_n) R_Greaterthanorequalto such that (with c R): (1) vol(e_1, ..., e_n) = 1 (2) Vol(upsilon_1, ..., Cupsilon_i, ..., V_n) = |c| Vol(upsilon_1, ...,upsilon_n) (3) VOl(upsilon_1, ...,upsilon_i + Cupsilonj, ..., upsilon_n) = Vol(upsilon_1, ..., upsilon_i,..., upsilon_n) (here upsilon_i + cupsilon_j is in the i-th spot, with i notequalto j) (4) vol(upsilon_1, ...,upsilon_i, ...,upsilon_j, ...,upsilon_n) = vol(upsilon_1, ...,upsilon_j, ..., upsilon_i, ...,upsilon_n) (here upsilon_i and upsilon_j are in the i-th and j-th spots on the LHS, and swap on the RHS) Using these axioms, prove that |det(upsilon_1, ..., upsilon_n)| = vol(upsilon_1, ..., upsilon_n).

Solution

The axioms (1) to (4) are exactly the properties that determine (uniquely) the determinant function (except for 2, where c instead of |c| is required).

In other words, determinant is the unique multilinear form satsifying the axioms (1) to (4) , again using c instead of |c| in (2)).

As volume is always positive and the volume of the paralleopiped formed by the unit vectors in the standard basis is 1 , the result follows. ( by taking the modulus of the determinant)