link to my thesis

A.3. EXAMPLE 251
The first step of the linearization technique, transforms the problem M(µ, λ) v =0
into an equivalent problem where the eigenvector z is structured as:
⎛
z = ⎝
v
µv
µ 2 v
⎞
⎠ ,
(A.19)
and where the involved matrices are linear in µ:
⎛⎛
⎞ ⎛
0 I 0
−I 0 0
⎜⎜
⎟ ⎜
⎝⎝
0 0 I ⎠ + µ ⎝ 0 −I 0
M 0 (λ) M 1 (λ) M 2 (λ)
0 0 M 3 (λ)
This equals:
⎛⎛
⎜⎜
⎝⎝
⎞⎞
⎛
⎟⎟
⎜
⎠⎠ z = ⎝
0 I 0
0 0 I
(N 0 + λN 2 + λ 2 N 4 + λ 3 N 9 ) (N 1 + λN 5 + λ 2 N 7 ) (N 3 + λN 8 )
0
0
0
⎞
⎞
⎟
⎠ .
⎟
⎠ +
(A.20)
⎛
⎞⎞
⎛ ⎞
−I 0 0
0
⎜
⎟⎟
⎜ ⎟
µ ⎝ 0 −I 0 ⎠⎠ z = ⎝ 0 ⎠ .
0 0 N 6 0
(A.21)
Furthermore, the blocks I and 0 denote identity blocks and zero blocks, respectively,
of dimensions n × n. The dimension of the matrices involved in the problems (A.20)
and (A.21) is therefore 3n × 3n.
In the second step, the linearization with respect to λ is carried out. First the
matrix involved in Equation (A.21) is written as:
where
K 0 = ⎝
(K 0 + K 1 λ + K 2 λ 2 + K 3 λ 3 )+µ(L 0 + L 1 λ + L 2 λ 2 + L 3 λ 3 ) (A.22)
⎛
0 I 0
⎞
⎛
0 0 I ⎠ , K 1 = ⎝
0 0 0
⎞
0 0 0 ⎠ ,
⎛
K 2 = ⎝
0 0 0
⎞
⎛
0 0 0 ⎠ , K 3 = ⎝
0 0 0
0 0 0
N 9 0 0
⎞
⎠ ,
(A.23)
⎛
L 0 = ⎝
−I 0 0
⎞
⎛
0 −I 0 ⎠ , and L 1 = L 2 = L 3 = ⎝
0 0 0
0 0 0
0 0 0
⎞
⎠ .
Here the blocks I and 0 denote again identity and zero blocks of dimensions n × n.