## 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.