## 5.13. Determinants and Independence
of Vectors

### Theorem 5.13.

A set of *n* vectors
{
A
1
,⋯,
A
n
}
in *n*-space
is independent
iff
d(
A
1
,⋯,
A
n
)≠0
.

#### Proof

The negation of Theorem 5.8, which says
dependence implies
d(
A
1
,⋯,
A
n
)=0
,
gives
d(
A
1
,⋯,
A
n
)≠0
implies
{
A
1
,⋯,
A
n
}
is independent,
thus proving the “if” part of the theorem.

To prove the “only
if” part, let *V* be the linear *n*-space in question. Since
{
A
1
,⋯,
A
n
}
is independent,
it forms a basis for *V*. By Theorem 4.12, there exists a linear
transformation
T:V→V
such that

T(
A
k
)=
I
k
for
k=1,⋯,n

where
I
k
is the *k*th
unit coordinate vector. In terms of
components in the *I*_{j}
basis, we have

∑
j=1
n
t
ij
(
A
k
)
j
=
(
I
k
)
i

Setting
B=(
b
ij
=
t
ji
)
,
this becomes

A
k
B=
I
k
for
k=1,⋯,n

where *A*_{k} and *I*_{k} are row matrices.
By Lemma 5.10, we have
A
k
B=
(
AB
)
k
,
where *A* is a matrix with rows
A
1
,⋯,
A
n
. Hence,

(
AB
)
k
=
I
k
for
k=1,⋯,n

which means
AB=I
,
i.e., *A* is nonsingular and by Theorem
5.12,
detA≠0
. QED.