## 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:VV  such that

T( A k )= I k                for k=1,,n

where I k  is the kth unit coordinate vector.  In terms of components in the Ij 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 Ak and Ik 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, detA0 .  QED.