2.5 Linear Independence and Linear Dependence
The concepts of linear independence and linear dependence are very useful when working with vectors. A
collection of vectors is linearly dependent if one of them can be expressed as a linear combination of the others.
Otherwise they are linearly dependent. In most books this is usually stated in the following symmetric form.
Definition 1. Vectors v1, …, vn are linearly independent if the only way to write the zero vector as a linear
combination of them is with all coefficients equal to zero, i.e. if
(1)
c1v1 + c2v2 + … + cnvn = 0
for some scalars c1, c2, …, cn implies c1 = 0, c2 = 0, …, cn = 0. Vectors v1, …, vn are linearly dependent if there
exist scalars c1, c2, …, cn not all zero such that (1) holds.
c1
c2
Remark. If A is the matrix whose columns are v1, …, vn, then c1v1 + c2v2 + … + cnvn = A . . So v1, …, vn are
 
cn
linearly independent if and only if the nullspace of A consists of only the zero vector. This occurs if and only if
when we put A in echelon form every variable is a pivot variable, i.e. there are no columns corresponding to free
variables, i.e. each column has a row with a leading 1. This occurs if and only if the solution x to Ax = b is
unique, provided there exists a solution.
1
2
1
2
0
Example 1. The vectors v1 = 2 and v2 = 4 are linearly dependent because (-2) 2 + 4 = 0.
2
7
8
0
-3
Example 2. Are the vectors v1 =  , v2 =   and v3 = -1 linearly independent or linearly dependent?
0
0
6
2
7
 
 
 8  0
0
-3
Answer. The equation (1) becomes c1  + c2  + c3-1 = 0 which in turn is equivalent to the system
0
0
 6  0
2c1 + 7c2 + 8c3 = 0
- 3c2 - c3 = 0 . The last equation implies c3 = 0. Substituting into the second equation gives c2 = 0.
6c3 = 0
Substituting into the first equation gives c1 = 0. So the vectors v1, v2 and v3 are independent.
As mentioned above, another way to see that a collection of vectors is dependent is if one can be written as a
superposition of the others.
Proposition 1. v1, …, vn are linearly dependent if one can be written as a superposition of the others.
Proof. Suppose vj = c1v1 + c2v2 + … + cj-1vj-1. Then c1v1 + … + cj-1vj-1 + (-1)vj + 0vj+1 + … + 0vn = 0 so
v1, …, vn are linearly dependent. Conversely, suppose v1, …, vn are linearly dependent. Then (1) holds and not
all the ck are zero. Let j be the largest index such that cj  0. Then c1v1 + c2v2 + … + cjvj = 0 and vj = - Error!
c1v1 + c2v2 + … + cj-1vj-1.
2.5 - 1