in
ℝ
3
ℒ
=
(
12
-18
-8
5
-8
-3
11
-15
-8
)
p
(
x
)
=
det
(
x
ℐ
3
-
ℒ
)
=
2
+
5
⁢
x
+
4
⁢
x
2
+
x
3
=
(
x
+
2
)
⁢
(
x
+
1
)
2
Eigenvalue
λ
1
=
-2
(
-14
18
8
0
-5
6
3
0
-11
15
6
0
)
→
(
1
0
-1
0
0
3
-1
0
0
0
0
0
)
Eigenvalue
λ
2
=
-1
(
-13
18
8
0
-5
7
3
0
-11
15
7
0
)
→
(
1
0
-2
0
0
1
-1
0
0
0
0
0
)