5. “Standard”
R
n
, the set of real
[
n
, 1]
matrices
X
(i.e. column
n
vectors), is likewise
the first example of a
real vector space
, i.e a vector space taken with real scalars.
Subscribe to view the full document.
8
CHAPTER 1.
LINEAR ALGEBRA (APPROX. 9 LECTURES)
The central theme here is that many important examples exist which are not naturally
a
C
n
or
R
n
.
Question:
Why is
C
1
not the same as
R
2
?
Answer:
The “set” is the same, namely pairs (
a
,
b
) of real numbers, but the scalars are
different!
Question:
Why is
V
=
C
[
a
,
b
], the set of continuous functions from the interval [
a
,
b
] to
C
, a complex vector space?
Answer:
To check closure under both addition and scalar multiplication note: if
f
,
g
are
continuous functions then (
αf
+
βg
)(
x
) =
αf
(
x
) +
βg
(
x
) for all
x
defines a continuous
function “
αf
+
βg
” for any complex numbers
α
,
β
. From this we can easily check that
C
[0, 1] satisfies the remaining algebraic properties of a vector space.
Idea:
Think of the
x
in
f
(
x
) as a continuous analogue of the
i
on
a
i
where
A
= (
a
i
)
n
i
=1
.
In some sense
f
is an
n
–vector in the case
n
=
∞
.
Notation:
V
=
C
[
a
,
b
] denotes the vector space of continuous functions on the interval
[
a
,
b
].
Idea:
: Many vector spaces
W
arise as subsets of a larger vector space
V
.
Definition 3.
A subset
W
of a vector space
V
is called a
subspace
if it is
1. closed under addition;
2. closed under scalar multiplication.
Every subspace is itself a vector space.
The point of this is that all the vector space properties for a subset
W
∈
V
follow if
we check only the two closure conditions at once.
Examples 2.
1.
V
=
R
3
.
The full list of all subspaces of
R
3
consists of:
{
0
}
; all
possible lines containing
0
; all possible planes containing
0
; and finally
R
3
itself.
2.
R
+
=
{
x
≥
0
} ⊂
R
. This is closed under addition but
NOT
closed under scalar
multiplication, and is
NOT
a subspace.
3. Linear ODE with parameter
n
:
L
[
y
] =
x
2
y
+
xy
+ (
x
2

n
2
)
y
= 0,
for
x >
0.
(1.1)
The solutions
y
(
x
)
of (1.1) are called Bessel functions, and form a subspace of
C
(0,
∞
)
. To check this note that
L
[
αy
1
+
βy
2
] =
α
L
[
y
1
] +
β
L
[
y
2
]
, so if
y
1
,
y
2
solve
the ODE
(1.1)
, then
αy
1
+
βy
2
also solves the ODE.
Definition 4. Span
Given
x
1
,
. . . x
m
∈
V
define
span
(
x
1
,
. . . x
m
) =
set of all linear combinations
m
n
=1
α
n
x
n
.
1.3.
BASIS
9
Fact:
span
is always a subspace!
Example 1.
It will turn out that every solution of (1.1) is of the form
y
(
x
) =
α
1
J
n
(
x
) +
α
2
Y
n
(
x
)
where
J
n
is the Bessel function of the first kind of order
n
and
Y
n
is the Bessel
function of the second kind of order
n
. Therefore, the solution space of (1.1) is
span
(
J
n
,
Y
n
)
.
Typical problem in
V
=
C
n
(or
R
n
)
: how do you check if a given
n
–vector
X
lies in
the
span
(
u
1
,
. . .
u
m
)
⊂
C
n
?
Idea:
Build a
matrix
U
out of the vectors
u
1
,
. . .
u
m
:
U
= [
u
1
,
u
2
,
. . .
,
u
m
]
(1.2)
U
is a matrix with
m
columns and
n
rows.
Then:
X
∈
span
(
u
1
,
. . .
u
m
)
⇔
X
=
m
i
=1
α
i
u
i
for some
a
= (
α
1
,
. . .
,
α
m
)
T
⇔
X
=
Ua
for some
a
= (
α
1
,
. . .
,
α
m
)
T
⇔
the augmented matrix
U X
is consistent.
Subscribe to view the full document.
 Winter '10
 kovarik