Graphics Reference
In-Depth Information
Expanding (
9.7
) using the top row we have
λ
+
1
=
1
−
λ
0
01
−
λ
−
0
0
−
λ
−
10
λ
−
λ
2
+
−
λ
+
1
−
λ
=
0
λ
2
−
λ
3
+
1
−
λ
=
0
−
λ
3
+
λ
2
−
λ
+
1
=
0
λ
3
−
λ
2
+
λ
=
1
.
When working with 3
3 matrices we always end up with a cubic in
λ
,forwhich
there can be three types of solution:
×
1. One real and two complex conjugate solutions.
2. Three real solutions including the possibility of a double solution.
3. Three distinct real solutions.
It is clear that
λ
=
1 is one such real root, which satisfies our requirement for an
eigenvalue. We could also show that the other two roots are complex conjugates.
Substituting
λ
=
1 in the original equations associated with (
9.7
) to reveal the
eigenvector, we have
⎧
⎨
−
+
+
=
x
0
y
z
0
0
x
+
0
y
+
0
z
=
0
⎩
−
x
+
0
y
−
z
=
0
.
=
=
It is obvious from the 1st and 3rd equations that
x
0. However, all three
equations multiply the
y
term by zero, which implies that the associated eigenvector
is of the form
z
T
, which is the
y
-axis, as anticipated. Now let's find the
[
0
k
0
]
angle of rotation.
Using one of the above rotation matrices
R
β,y
and the trace operation:
⎡
⎤
cos
β
0 in
β
01
⎣
⎦
R
β,y
=
0
−
sin
β
0
cos
β
Tr
(
R
β,y
)
=
1
+
2 cos
β
therefore,
β
=
arccos
((
Tr
(
R
β,y
)
−
1
)/
2
).
To illustrate this, let
β
=
90°:
⎡
⎤
001
010
⎣
⎦
R
90
°
,y
=
−
100
=
Tr
(
R
90
°
,y
)
1
therefore,
β
=
arccos
((
1
−
1
) /
2
)
=
90°
.