Civil Engineering Reference
In-Depth Information
Table 2.19.
Example 2.16 Faddev-Leverrier method
[B
1
]=[A] =
1
2
3
1
2
3
−10
0
2
−10
0
2
−2
4
8
−2
4
8
[B
2
]=([A]([B
1
]−p
1
I)
p
1
=tr[B
1
] = 9
−
1
2
3
*
1
2
3
9
−10
0
2
−10
0
2
9
−2
4
8
−2
4
8
9
1
2
3
*
−8
2
3
=
−34
−4
4
−10
0
2
−10
−9
2
76
−12
−32
−2
4
8
−2
4
−1
−40
−8
−6
[B
3
]=([A]([B
2
]−p
2
I)
p
2
=(1/2)tr[B
2
] = −26
−
1
2
3
*
−34
−4
4
−26
−10
0
2
76
−12
−32
−26
−2
4
8
−40
−8
−6
−26
1
2
3
*
−8
−4
4
=
24
0
0
−10
0
2
76
14
−32
0
24
0
−2
4
8
−40
−8
20
0
0
24
p3=(1/3)tr[B
3
] = 24
a.
Divide all the
x
's by first
x
value.
b.
Divide all the
x
's by the largest
x
.
c.
Normalize to a unit length.
4.
Use the components of the normalized vector as improved values
of
x
.
5.
Repeat steps 2 through 4 until the previous values differ from the
new values by less than some small value (
e
).
