5.7. Uniqueness of the Determinant Function

Eq(5.8) showed that axioms 1 and 2 alone can be used to define a determinant function.  This raises the question of uniqueness of such functions.

Theorem 5.6.  Uniqueness Theorem For Determinants

Let d be a function satisfying all 3 axioms for a determinant function of order n.  If f is another function satisfying axioms 1 and 2, then for every n ´ n matrix A, we have

        f( A )=d( A )f( I )                          (5.9)

Hence, if f also satisfies axioms 3, then f( A )=d( A )  so that d is unique.

Proof

Consider

        g( A )=f( A )d( A )f( I )                      (a)

Since both f and d satisfy axioms 1 and 2, so does g.  However, using d( I )=1 , we have

        g( I )=f( I )d( I )f( I )=0

so that g does not satisfy axiom 3.  Thus, eq(5.8) becomes

        g( A )=c( A )g( I )=0

Putting this into eq(a), we recover eq(5.9).  QED.