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Then
=
4
−
λ
1
14
0
−
λ
λ
)
2
(
4
−
−
1
=
0
λ
2
−
8
λ
+
16
−
1
=
0
λ
2
−
8
λ
+
15
=
0
(λ
−
5
)(λ
−
3
)
=
0
.
Thus
λ
3, are the two eigenvalues we observed in Fig.
4.1
.Next,we
substitute the two values of
λ
in
4
=
5 and
λ
=
x
y
0
0
−
λ
1
=
−
14
λ
=
to extract the eigenvectors. Let's start with
λ
5:
−
x
y
0
0
11
1
=
−
1
T
. Next, we substitute
which represents the equation
y
=
x
or the vector
[
kk
]
=
λ
3:
11
11
x
y
0
0
=
T
.
Thus we have discovered that the transform possesses two eigenvectors
which represents the equation
y
=−
x
or the vector
[−
kk
]
T
[
kk
]
T
[−
]
=
=
and
3, as predicted.
The characteristic equation may have real or complex solutions, and if they are
complex, there are no real eigenvectors. For example, we have already come across
the 2D transform for rotating points about the origin:
kk
and their respective eigenvalues
λ
5 and
λ
cos
β
−
sin
β
A
=
sin
β
cos
β
and we would not expect this to have any real eigenvectors, as this would imply that
it shows a rotational preference to certain points. Let's explore this transform to see
how the characteristic equation behaves.
The characteristic equation is
=
cos
β
−
λ
−
sin
β
0
sin
β
cos
β
−
λ
where
β
is the angle of rotation. Therefore,
λ)
2
sin
2
β
(
cos
β
−
+
=
0
λ
2
cos
2
β
sin
2
β
−
+
+
=
2
λ
cos
β
0
λ
2
−
2
λ
cos
β
+
1
=
0
.
