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LINEAR INDEPENDENCE
 Definition: An indexed set of vectors {v1, …, vp} in
is said to be linearly independent if the vector
equation
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
----(1)
1 1 2 2
v v ... v 0
p p
x x x
+ + + =
1 1 2 2
v v ... v 0
p p
c c c
+ + + =
LINEAR INDEPENDENCE
 Equation (1) 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.
 Example 1: Let , , and .
1
1
v 2
3
 
 
=
 
 
 
2
4
v 5
6
 
 
=
 
 
 
3
2
v 1
0
 
 
=
 
 
 
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 following equation.
1 2 3
1 4 2 0
2 5 1 0
3 6 0 0
x x x
       
       
+ + =
       
       
       
LINEAR INDEPENDENCE
 Row operations on the associated augmented matrix
show that
.
 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.
1 4 2 0 1 4 2 0
2 5 1 0 0 3 3 0
3 6 0 0 0 0 0 0
   
   
− −
   
   
   

LINEAR INDEPENDENCE
b. To find a linear dependence relation among v1,
v2, and v3, row reduce the augmented matrix
and write the new system:
 Thus, , , and x3 is free.
 Choose any nonzero value for x3—say, .
 Then and .
1 0 2 0
0 1 1 0
0 0 0 0
−
 
 
 
 
 
1 3
2 3
2 0
0
0 0
x x
x x
− =
+ =
=
1 3
2
x x
= 2 3
x x
= −
3
5
x =
1
10
x = 2
5
x = −
LINEAR INDEPENDENCE
 Substitute these values into equation (1) and obtain
the equation below.
 This is one (out of infinitely many) possible linear
dependence relations among v1, v2, and v3.
1 2 3
10v 5v 5v 0
− + =
LINEAR INDEPENDENCE OF MATRIX COLUMNS
 Suppose that we begin with a matrix
instead of a set of vectors.
 The matrix equation can be written as
.
 Each linear dependence relation among the columns of
A corresponds to a nontrivial solution of .
 Thus, the columns of matrix A are linearly independent if
and only if the equation has only the trivial
solution.
x 0
A =
1 1 2 2
a a ... a 0
n n
x x x
+ + + =
x 0
A =
x 0
A =
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 has only
the trivial solution when .
 The zero vector is linearly dependent because
has many nontrivial solutions.
1
v 0
x =
v 0
≠
1
0 0
x =
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.
Therefore they are linearly dependent if they lie on
the same line.
 The set is linearly independent if and only if neither
of the vectors is a multiple of the other, i.e. if they
span a plane.
SETS OF TWO OR MORE VECTORS
 Theorem 7: Characterization of Linearly Dependent
Sets
 An indexed set 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 , then
some vj (with ) is a linear combination of the
preceding vectors, v1, …, .
1
{v ,...,v }
p
S =
1
v 0
≠
1
j >
1
vj−
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.
 Example 2: Let and . Describe the
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.
3
u 1
0
 
 
=
 
 
 
1
v 6
0
 
 
=
 
 
 
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 .
 Span {u, v} is the x1x2-plane (with ).
 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 ).
 That vector must be w, since v is not a multiple of u.
3
0
x =
u 0
≠
SETS OF TWO OR MORE VECTORS
 So w is in Span {u, v}. See the figures given below.
 Example 2 generalizes to any set {u, v, w} in 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.
3

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 is
linearly dependent if .
 Proof: Let .
 Then A is , and the equation
corresponds to a system of n equations in p
unknowns.
 If , there are more variables than equations, so
there must be a free variable.
p n
>
n p
× x 0
A =
p n
>

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2 linear independence

  • 1. LINEAR INDEPENDENCE  Definition: An indexed set of vectors {v1, …, vp} in is said to be linearly independent if the vector equation 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 ----(1) 1 1 2 2 v v ... v 0 p p x x x + + + = 1 1 2 2 v v ... v 0 p p c c c + + + =
  • 2. LINEAR INDEPENDENCE  Equation (1) 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.  Example 1: Let , , and . 1 1 v 2 3     =       2 4 v 5 6     =       3 2 v 1 0     =      
  • 3. 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 following equation. 1 2 3 1 4 2 0 2 5 1 0 3 6 0 0 x x x                 + + =                        
  • 4. LINEAR INDEPENDENCE  Row operations on the associated augmented matrix show that .  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. 1 4 2 0 1 4 2 0 2 5 1 0 0 3 3 0 3 6 0 0 0 0 0 0         − −             
  • 5. LINEAR INDEPENDENCE b. To find a linear dependence relation among v1, v2, and v3, row reduce the augmented matrix and write the new system:  Thus, , , and x3 is free.  Choose any nonzero value for x3—say, .  Then and . 1 0 2 0 0 1 1 0 0 0 0 0 −           1 3 2 3 2 0 0 0 0 x x x x − = + = = 1 3 2 x x = 2 3 x x = − 3 5 x = 1 10 x = 2 5 x = −
  • 6. LINEAR INDEPENDENCE  Substitute these values into equation (1) and obtain the equation below.  This is one (out of infinitely many) possible linear dependence relations among v1, v2, and v3. 1 2 3 10v 5v 5v 0 − + =
  • 7. LINEAR INDEPENDENCE OF MATRIX COLUMNS  Suppose that we begin with a matrix instead of a set of vectors.  The matrix equation can be written as .  Each linear dependence relation among the columns of A corresponds to a nontrivial solution of .  Thus, the columns of matrix A are linearly independent if and only if the equation has only the trivial solution. x 0 A = 1 1 2 2 a a ... a 0 n n x x x + + + = x 0 A = x 0 A =
  • 8. 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 has only the trivial solution when .  The zero vector is linearly dependent because has many nontrivial solutions. 1 v 0 x = v 0 ≠ 1 0 0 x =
  • 9. 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. Therefore they are linearly dependent if they lie on the same line.  The set is linearly independent if and only if neither of the vectors is a multiple of the other, i.e. if they span a plane.
  • 10. SETS OF TWO OR MORE VECTORS  Theorem 7: Characterization of Linearly Dependent Sets  An indexed set 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 , then some vj (with ) is a linear combination of the preceding vectors, v1, …, . 1 {v ,...,v } p S = 1 v 0 ≠ 1 j > 1 vj−
  • 11. 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.  Example 2: Let and . Describe the 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. 3 u 1 0     =       1 v 6 0     =      
  • 12. 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 .  Span {u, v} is the x1x2-plane (with ).  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 ).  That vector must be w, since v is not a multiple of u. 3 0 x = u 0 ≠
  • 13. SETS OF TWO OR MORE VECTORS  So w is in Span {u, v}. See the figures given below.  Example 2 generalizes to any set {u, v, w} in 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. 3 
  • 14. 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 is linearly dependent if .  Proof: Let .  Then A is , and the equation corresponds to a system of n equations in p unknowns.  If , there are more variables than equations, so there must be a free variable. p n > n p × x 0 A = p n >