in
ℝ
3
ℒ
=
(
-7
6
4
-3
2
2
-3
3
1
)
p
(
x
)
=
det
(
x
ℐ
3
-
ℒ
)
=
2
+
5
⁢
x
+
4
⁢
x
2
+
x
3
=
(
x
+
2
)
⁢
(
x
+
1
)
2
Eigenvalue
λ
1
=
-2
(
5
-6
-4
0
3
-4
-2
0
3
-3
-3
0
)
→
(
1
0
-2
0
0
1
-1
0
0
0
0
0
)
Eigenvalue
λ
2
=
-1
(
6
-6
-4
0
3
-3
-2
0
3
-3
-2
0
)
→
(
3
-3
-2
0
0
0
0
0
0
0
0
0
)