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° .
Search WWH ::




Custom Search