Graphics Reference
In-Depth Information
This quadratic in
λ
is solved using
λ
=
−
b
±
√
b
2
−
4
ac
2
a
where
a
=
1,
b
=−
2 cos
β
,
c
=
1:
4 cos
2
β
2 cos
β
±
−
4
λ
=
2
cos
2
β
=
±
−
cos
β
1
sin
2
β
=
cos
β
±
−
λ
1
=
cos
β
+
i
sin
β
λ
2
=
cos
β
−
i
sin
β
which are complex numbers.
The corresponding complex eigenvectors are
1
i
v
1
=
1
−
.
v
2
=
i
Now let's investigate the eigenvectors associated with a 3D transform. We start
with the arbitrary transform
⎡
⎤
⎦
301
⎣
=
−
133
103
A
and its characteristic equation is
3
−
λ
0
1
−
13
−
λ
3
=
0
.
1
0
3
−
λ
Expanding the determinant using the top row we have
−
λ)
−
=
3
−
λ
3
03
−
13
−
λ
(
3
0
+
0
−
λ
1
0
λ)
2
(
3
−
λ)(
3
−
−
(
3
−
λ)
=
0
λ)
(
3
1
=
λ)
2
(
3
−
−
−
0
λ)
λ
2
8
=
(
3
−
−
6
λ
+
0
−
−
−
=
(
3
λ)(λ
4
)(λ
2
)
0
which has solutions
λ
=
2, 3, 4. Let's substitute these values of
λ
in the original
equations to reveal the eigenvectors:
⎧
⎨
(
3
−
λ)x
+
z
=
0
−
x
+
(
3
−
λ)y
+
3
z
=
0
⎩
x
+
(
3
−
λ)z
=
0
.
