Idea:
The trick of writing
m
i
=1
α
i
u
i
=
Ua
(1.3)
for
a
= (
α
1
,
. . .
,
α
m
)
T
is very helpful!
1.3
Basis
Given a set
{
u
1
,
. . .
,
u
m
}
of vectors in
V
, the span
W
=
span
(
u
1
,
. . .
,
u
m
) is always a
subspace of
V
. This means every
x
∈
W
can be written as
x
=
m
n
=1
α
n
u
n
(1.4)
for some scalars
α
1
,
α
m
.
However, some vector, say
u
i
, may be
redundant
in the sense
that
u
i
is itself a linear combination of the remaining vectors in
{
u
1
,
. . .
,
u
m
}
. When this
is the case,
{
u
1
,
. . .
,
u
m
}
is said to be
linearly dependent
.
We want to focus on sets
{
u
1
,
. . .
,
u
m
}
that have no such redundancy. Such a set of
vectors is called
linearly independent
; formally, this means that the only set of scalars
α
i
,
i
= 1 :
m
such that
m
n
=1
α
n
u
n
= 0
is
α
i
= 0, for all
i
.
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10
CHAPTER 1.
LINEAR ALGEBRA (APPROX. 9 LECTURES)
Definition 5.
A set
{
u
1
,
. . .
,
u
m
}
of vectors in
W
forms a
basis for
W
iff these two
properties are true:
1.
span
(
u
1
,
. . .
,
u
m
) =
W
2.
{
u
1
,
. . .
,
u
m
}
is
linearly independent
.
In other words, “
{
u
1
,
. . .
,
u
m
}
is a basis for
W
” means
x
=
∑
m
n
=1
α
n
u
n
is uniquely
solvable for any
x
∈
W
.
Examples 3.
1.
Standard basis
for both for
R
3
and
C
3
:
e
1
= (1, 0, 0)
T
,
e
2
= (0, 1, 0)
T
,
e
3
= (0, 0, 1)
T
.
Check this is a basis: For any
X
= (
x
1
,
x
2
,
x
3
)
T
,
X
=
∑
i
α
i
e
i
is uniquely solved by
x
i
=
α
i
,
i
= 1, 2, 3
.
Note that in case
V
=
C
3
(respectively
R
3
), the scalars are
complex (real).
2.
u
1
= (1, 1, 0)
T
,
u
2
= (1,

1, 0)
T
,
u
3
= (1, 1, 1)
T
is another basis for
R
3
and
C
3
. Why?
Answer by row reduction: For any
X
, to find
A
= (
α
1
,
α
2
,
α
3
)
T
such that
X
= (
x
1
,
x
2
,
x
3
)
T
=
α
1
u
1
+
α
2
u
2
+
α
3
u
3
we use the same trick as before. We write
U
= [
u
1
,
u
2
,
u
3
]
and solve
X
=
UA
for the
3
unknown components of
A
. By row reducing the augmented matrix
[
U

X
]
,
we can show the system is uniquely solvable for any
X
.
More succinctly, for any
X
∈
C
3
, its coordinates
A
∈
C
3
are computed by
A
=
U

1
X
where
U

1
is the
inverse of
U
.
Algorithm 1.
(Computing
U

1
): use GaussJordan elimination on the fully aug
mented
[3, 6]
matrix
[
U

I
]
→
[
I

U

1
]
Exercise:
Review Algorithm 1 and compute the inverse of
U
=
1
1
1
1

1
1
0
0
1
.
3.
V
=
C
[0, 1]: The three functions
{
cos
x
, sin
x
, 1
}
are linearly independent, which
means they form a basis for their span.
On the contrary, the three functions
{
cos
2
x
, sin
2
x
, 1
}
are linearly dependent because of the identity
1
·
cos
2
x
+ 1
·
sin
2
x
+ (

1)
·
1 = 0
(1.5)
4. The functions
v
n
(
x
) =
e
inx
for
n
=

N
:
N
form a basis for the vector space
W
N
.
1.4.
LINEAR OPERATORS
11
5. The pair of functions
J
n
,
Y
n
form a basis for the solution space of (1.1). We some
times say they form a
fundamental set of solutions
.
Definition 6.
1.
V
has dimension
N
means it has a set of
N
linearly independent
vectors, but no set of
N
+ 1
linearly independent vectors
2.
V
is
infinite dimensional
means it has a set of
N
linearly independent vectors for
every
N
.
Theorem 1.
(“Dimension theorem”) Any two bases
{
u
1
,
. . .
,
u
m
}
and
{
v
1
,
. . .
,
v
n
}
for a
vector space
V
with finite dimension
N
have the same number of vectors
m
=
n
=
N
.
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 Winter '10
 kovarik