Examples 6.
1. Rotation
R
by
θ
about the
e
3
=
z
–axis; choose the standard basis for
R
3
R
e
1
=
cos
θ
e
1
+ sin
θ
e
2
=
r
11
e
1
+
r
21
e
2
+
r
31
e
3
R
e
2
=

sin
θ
e
1
+ cos
θ
e
2
=
r
12
e
1
+
r
22
e
2
+
r
32
e
3
R
e
3
=
e
3
=
r
13
e
1
+
r
23
e
2
+
r
33
e
3
Reading off gives the matrix (note the transpose!):
R
=
R
z
,
θ
=
cos
θ

sin
θ
0
sin
θ
cos
θ
0
0
0
1
(1.7)
2. Similarly, rotations by
θ
about the
x
and
y
axes are
R
=
R
x
,
θ
=
1
0
0
0
cos
θ

sin
θ
0
sin
θ
cos
θ
,
R
=
R
y
,
θ
=
cos
θ
0

sin
θ
0
1
0
sin
θ
0
cos
θ
(1.8)
3. For
V
= polynomials of degree 3
, choose basis
{
e
1
,
e
2
,
e
3
,
e
4
}
=
{
1,
x
,
x
2
,
x
3
}
Take
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14
CHAPTER 1.
LINEAR ALGEBRA (APPROX. 9 LECTURES)
linear operator
D
=
d
dx
:
V
→
V
.
D
e
1
=
0 =
4
i
=1
d
i
1
e
i
D
e
2
=
1 =
4
i
=1
d
i
2
e
i
D
e
3
=
2
x
=
4
i
=1
d
i
3
e
i
D
e
4
=
3
x
2
=
4
i
=1
d
i
4
e
i
From the first equation we can read off the first
column
of
D
:
d
11
=
d
21
=
d
31
=
d
41
= 0
Continuing, gives the matrix
D
= [
d
ij
] =
0
1
0
0
0
0
2
0
0
0
0
3
0
0
0
0
One can summarize these steps in a memorable way:
D
[1,
x
,
x
2
,
x
3
] = [0, 1, 2
x
, 3
x
3
] = [1,
x
,
x
2
,
x
3
]
D
In other words,
DU
=
U
D
.
Now check that
D
does the job.
For example,
P
(
x
) = 1 + 2
x
3
has coordinates
(1, 0, 0, 2)
T
. Apply
D
to this column vector:
0
1
0
0
0
0
2
0
0
0
0
3
0
0
0
0
1
0
0
2
=
0
0
6
0
which gives the coordinates for
P
(
x
) = 6
x
2
.
Question:
What happens when we compose linear operators?
Answer:
Let
A
and
B
be linear operators
V
→
V
: then
C
=
B ◦ A
:
V
→
V
is also a
linear operator. Given a basis
{
u
i
}
n
i
=1
for
V
: how do their matrices
A
,
B
,
C
relate? Well,
C
u
i
=
B
(
A
u
i
) =
B
(
m
j
=1
a
ji
u
j
) =
m
j
=1
a
ji
(
B
u
j
)
=
m
j
=1
a
ji
(
p
k
=1
b
kj
u
k
) =
p
k
=1
(
m
j
=1
a
ji
b
kj
)
u
k
=
p
k
=1
c
ki
u
k
1.4.
LINEAR OPERATORS
15
from which we find matrix multiplication again:
c
ki
=
m
j
=1
b
kj
a
ji
or
C
=
BA
Examples 7.
1. Figure 1.1 shows several compositions of linear operators. The impor
tant fact about the matrix
A
of
A
given the basis
{
u
i
}
is that the defining equations
(1.6)
are equivalent to
A ◦ U
=
U ◦
A
and
A
=
U ◦
A
◦ U

1
.
(1.9)
2. The general rotation in
R
3
can be written
R
=
R
z
,
ψ
R
x
,
θ
R
z
,
φ
(1.10)
φ
,
θ
,
ψ
are called the three “Euler angles”. Note again that matrix multiplication is
not commutative!
3.
D
2
is the second derivative matrix.
Question:
Given
A
:
V
→
V
and two bases
{
u
i
}
and
{
v
i
}
for
V
, how does the
u
matrix
A
compare to the
v
matrix
A
?
Answer:
This is called
change of basis
.
In Assignment 2 you are asked to show that
A
=
C
u
,
v
A
C

1
u
,
v
where
C
u
,
v
=
V

1
U
converts
u
coordinates of any vector
b
∈
V
into its
v
coordinates.
Question:
How do we get a matrix for an operator
A
:
V
→
W
?
Answer:
We need two bases: a basis
{
u
i
}
i
=1:
N
for
V
and a basis
{
w
i
}
i
=1:
M
for
W
. Relative
to these two bases, the matrix of
A
has components
A
= (
a
ij
)
i
=1:
M
,
j
=1:
N
defined by the
equations
A
u
i
=
M
j
=1
a
ji
w
j
,
for all
i
= 1 :
N
Things go much the same as the special case when
W
=
V
and
{
u
i
}
=
{
w
i
}
. In summary,
one finds that equation (1.9) extends to
A ◦ U
=
W ◦
A
,
A
=
W ◦
A
◦ U

1
, where
U
is the
coordinate operator for the
u
basis and
W
is the coordinate operator for the
w
basis.
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16
CHAPTER 1.
LINEAR ALGEBRA (APPROX. 9 LECTURES)
1.5
Invertible Linear Operators
 Winter '10
 kovarik