Lecture 6: Linear Independence, Spanning, Basis and Dimension.
The homogeneous system
A
x
=
0
can be studied from a different perspective by
writing them as vector equations;
(1)
1
2

3
3
5
9
5
9
3
x
1
x
2
x
3
=
0
0
0
⇔
x
1
1
3
5
+
x
2
2
5
9
+
x
3

3
9
3
=
0
0
0
A set of vectors
{
v
1
, . . . ,
v
p
}
in
R
n
is said to be
linearly independent
if
x
1
v
1
+
· · ·
+
x
p
v
p
=
0
has only the trivial solution
x
1
=
· · ·
=
x
p
= 0. The set
{
v
1
, . . . ,
v
p
}
is said to be
linearly dependent
if there are weights
λ
1
, . . . , λ
p
not all 0, such that
λ
1
v
1
+
· · ·
+
λ
p
v
p
=
0
.
Ex 1
Determine if
v
1
=
1
3
5
,
v
2
=
2
5
9
,
v
3
=

3
9
3
are linearly independent?
Sol
They are linearly dependent if (1) has a nontrivial solution. Augmented:
1
2

3
0
3
5
9
0
5
9
3
0
∼
1
2

3
0
0

1
18
0
0

1
18
0
∼
1
2

3
0
0

1
18
0
0
0
0
0
∼
1
0
33
0
0

1
18
0
0
0
0
0
Since
x
3
is free there are nontrivial solutions
x
1
=

33
x
3
,
x
2
=18
x
3
,
x
3
is free. If we
e.g. let
x
3
=1 then
x
1
=

33 and
x
2
=18 so we have the linear dependence relation

33
1
3
5
+ 18
2
5
9
+ 1

3
9
3
=
0
0
0
⇔
1
2

3
3
5
9
5
9
3

33
18
1
=
0
0
0
The columns of
A
are linearly independent
⇔
A
x
=
0
has only the trivial solution.
Th
n
vectors
{
v
1
,
· · ·
,
v
n
}
in
R
m
are linearly dependent if
n > m
.
Pf
A
=[
v
1
· · ·
v
n
] is a
m
×
n
matrix. If
n>m
then the homogeneous system
A
x
=
0
has more unkonowns than equations so there must be free variables. Hence
A
x
=
0
has a nontrivial solution
x
6
=
0
so the columns of
A
are linearly dependent.
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 Spring '12
 DavidGlenn
 Linear Algebra, Algebra, Equations, Linear Independence, Vector Space, basis, Row echelon form, pivot columns

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