1 Linear Equations
in Linear Algebra
1.7
LINEAR INDEPENDENCE
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LINEAR INDEPENDENCE
Definition: An indexed set of vectors {v1, …, vp} in
ℝ𝑛𝑛 is said to be linearly independent if the vector
equation
x1v1 + x2 v 2 + ... + x p v p =
0
has only the trivial solution. The set {v1, …, vp} is
said to be linearly dependent if there exist weights
c1, …, cp, not all zero, such that
c1v1 + c2 v 2 + ... + c p v p =
0 (2)
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LINEAR INDEPENDENCE
Equation (2) is called a linear dependence relation
among v1, …, vp when the weights are not all zero.
An indexed set is linearly dependent if and only if it
is not linearly independent.
1 4 2
Example 1: Let v1 = 2 , v 2 = 5 , and v3 = 1 .
3 6 0
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LINEAR INDEPENDENCE
a. Determine if the set {v1, v2, v3} is linearly
independent.
b. If possible, find a linear dependence relation among
v1, v2, and v3.
Solution: We must determine if there is a nontrivial
solution of the equation on the previous slide.
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LINEAR INDEPENDENCE
Row operations on the associated augmented matrix
show that
1 4 2 0 1 4 2 0
2 5 1 0 ~ 0 −3 −3 0 .
3 6 0 0 0 0 0 0
x1 and x2 are basic variables, and x3 is free.
Each nonzero value of x3 determines a nontrivial
solution of (1).
Hence, v1, v2, v3 are linearly dependent.
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LINEAR INDEPENDENCE
b. To find a linear dependence relation among v1, v2,
and v3, completely row reduce the augmented
matrix and write the new system:
1 0 −2 0 x1 − 2 x3 =
0
0 1 1 0
0
x2 + x3 =
0 0 0 0 0=0
Thus, x1 = 2 x3 , x2 = − x3 , and x3 is free.
Choose any nonzero value for x3—say, x3 = 5.
Then x1 = 10 and x2 = −5 .
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LINEAR INDEPENDENCE
Substitute these values into equation (1) and obtain
the equation below.
10v1 − 5v 2 + 5v3 =
0
This is one (out of infinitely many) possible linear
dependence relations among v1, v2, and v3.
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LINEAR INDEPENDENCE OF MATRIX COLUMNS
Suppose that we begin with a matrix A = [ a1 a n ]
instead of a set of vectors.
The matrix equation Ax = 0 can be written as
x1a1 + x2a 2 + ... + xn a n =
0.
Each linear dependence relation among the columns of A
corresponds to a nontrivial solution of Ax = 0
The columns of matrix A are linearly independent if and
only if the equation Ax = 0 has only the trivial solution.
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SETS OF ONE OR TWO VECTORS
A set containing only one vector – say, v – is linearly
independent if and only if v is not the zero vector.
This is because the vector equation x1v = 0 has only
the trivial solution when v ≠ 0.
The zero vector is linearly dependent because x1 0 = 0
has many nontrivial solutions.
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SETS OF ONE OR TWO VECTORS
A set of two vectors {v1, v2} is linearly dependent if
at least one of the vectors is a multiple of the other.
The set is linearly independent if and only if neither
of the vectors is a multiple of the other.
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SETS OF TWO OR MORE VECTORS
THEOREM 7
Characterization of Linearly Dependent Sets
An indexed set S = {v1 ,..., v p } of two or more
vectors is linearly dependent if and only if at least one
of the vectors in S is a linear combination of the
others. In fact, if S is linearly dependent and v1 ≠ 0 ,
then some vj (with j > 1 ) is a linear combination of
the preceding vectors, v1, …,v j−1 .
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SETS OF TWO OR MORE VECTORS
Proof: If some vj in S equals a linear combination of
the other vectors, then vj can be subtracted from both
sides of the equation, producing a linear dependence
relation with a nonzero weight (−1) on vj.
[For instance, if= v1 c2 v 2 + c3 v3 , then
0 =(−1)v1 + c2 v 2 + c3 v3 + 0v 4 + ... + 0v p .]
Thus S is linearly dependent.
Conversely, suppose S is linearly dependent.
If v1 is zero, then it is a (trivial) linear combination of
the other vectors in S.
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SETS OF TWO OR MORE VECTORS
Otherwise, v1 ≠ 0 , and there exist weights c1, …, cp, not
all zero, such that
c1v1 + c2 v 2 + ... + c p v p =
0.
Let j be the largest subscript for which c j ≠ 0.
If j = 1 , then c1v1 = 0 , which is impossible because
v1 ≠ 0 .
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SETS OF TWO OR MORE VECTORS
So j > 1, and
c1v1 + ... + c j v j + 0v j + 0v j +1 + ... + 0v p =
0
cjv j =−c1v1 − ... − c j −1v j −1
c1 c j −1
v j = − v1 + ... + − v j −1.
cj cj
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SETS OF TWO OR MORE VECTORS
Theorem 7 does not say that every vector in a linearly
dependent set is a linear combination of the preceding
vectors.
A vector in a linearly dependent set may fail to be a
linear combination of the other vectors.
3 1
Example 4: Let u = 1 and v = 6 . Describe the
0 0
set spanned by u and v, and explain why a vector w is
in Span {u, v} if and only if {u, v, w} is linearly
dependent.
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SETS OF TWO OR MORE VECTORS
Solution: The vectors u and v are linearly
independent because neither vector is a multiple of
the other, and so they span a plane in ℝ3 .
Span {u, v} is the x1x2-plane (with x3 = 0).
If w is a linear combination of u and v, then {u, v, w}
is linearly dependent, by Theorem 7.
Conversely, suppose that {u, v, w} is linearly
dependent.
By theorem 7, some vector in {u, v, w} is a linear
combination of the preceding vectors (since u ≠ 0 ).
That vector must be w, since v is not a multiple of u.
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SETS OF TWO OR MORE VECTORS
So w is in Span {u, v}. Fig. 2 below
Example 4 generalizes to any set {u, v, w} in ℝ3 with
u and v linearly independent.
The set {u, v, w} will be linearly dependent if and
only if w is in the plane spanned by u and v.
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SETS OF TWO OR MORE VECTORS
THEOREM 8
If a set contains more vectors than there are entries in
each vector, then the set is linearly dependent. That is,
any set {v1, …, vp} in ℝ𝑛𝑛 is linearly dependent if p > n .
Proof: Let A = v1 v p .
Then A is n × p , and the equation Ax = 0 corresponds
to a system of n equations in p unknowns.
If p > n , there are more variables than equations, so
there must be a free variable.
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SETS OF TWO OR MORE VECTORS
Hence Ax = 0 has a nontrivial solution, and the
columns of A are linearly dependent.
See the figure below for a matrix version of this
theorem.
Theorem 8 says nothing about the case in which the
number of vectors in the set does not exceed the
number of entries in each vector.
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SETS OF TWO OR MORE VECTORS
THEOREM 9
If a set S = {v1 ,..., v p } in ℝ𝑛𝑛 contains the zero vector,
then the set is linearly dependent.
Proof: By renumbering the vectors, we may suppose
v1 = 0 .
Then the equation 1v1 + 0v 2 + ... + 0v p =
0 shows
that S in linearly dependent.
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