Proof (exercise):



(⟹) Suppose that v 1 , v 2 ,  , v n  is linearly dependent.
Then, by definition,




c1 v 1  c 2 v 2    c n v n  0
Without loss of generality, suppose
c1  0
has a nontrivial solution.
. Then




c1 v 1   c 2 v 2  c 3 v 3    c n v n
c3 
cn 
c2 

v1  
v2 
v3  
vn
c1
c1
c1
(since
c1  0
)
Thus, one of the vectors can be written as a linear combination if the others.
(⟸) Suppose that at least one of the vectors can be written as a linear combination of the
others.
Without loss of generality, suppose
Rearranging, we have



v1  k2v 2    kn v n




c1 v 1  c 2 v 2    c n v n  0
Since
c1  0
has the solution
, this solution is nontrivial.


k2 , k3,  , kn




1v 1  k 2 v 2    k n v n  0
Thus,

for some scalars
Therefore, v 1 , v 2 ,  , v n  is linearly dependent.
c1  1
,
ci  ki
for
i  2 , 3, , n
.
.