Page 205 - 35Linear Algebra

10.1 Showing Linear Dependence 205

1
2
3
has any solutions for c , c , c . We can rewrite this as a homogeneous system by
building a matrix whose columns are the vectors v 1 , v 2 and v 3 :
1
 
c

c
v 1 v 2 v 3  2  = 0.
c 3

This system has solutions if and only if the matrix M = v 1 v 2 v 3 is singular, so
we should find the determinant of M:
 
0 1 1
1 1
det M = det  0 2 2  = det = 0.
2 2
1 1 3
Therefore nontrivial solutions exist. At this point we know that the vectors are
linearly dependent. If we need to, we can find coefficients that demonstrate linear
dependence by solving

     
0 1 1 0 1 1 3 0 1 0 2 0
0 2 2 0 ∼ 0 1 1 0 ∼ 0 1 1 0 .
     
1 1 3 0 0 0 0 0 0 0 0 0

The solution set {µ(−2, −1, 1) | µ ∈ R} encodes the linear combinations equal to zero;
3
2
1
any choice of µ will produce coefficients c , c , c that satisfy the linear homogeneous
equation. In particular, µ = 1 corresponds to the equation
3
2
1
c v 1 + c v 2 + c v 3 = 0 ⇒ −2v 1 − v 2 + v 3 = 0.
Reading homework: problem 1

1
k
Definition Any sum of vectors v 1 , . . . , v k multiplied by scalars c , . . . , c ,
namely
k
1
c v 1 + · · · + c v k ,
is called a linear combination of v 1 , . . . , v k .

Theorem 10.1.1 (Linear Dependence). An ordered set of non-zero vectors
(v 1 , . . . , v n ) is linearly dependent if and only if one of the vectors v k is ex-
pressible as a linear combination of the preceding vectors.

Proof. The theorem is an if and only if statement, so there are two things to
show.

205