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248 APPENDIX A. LINEARIZING A 2-PARAMETER EIGENVALUE PROBLEM
Then the following holds:
M(µ, λ) v =
(
) (
)
M 0 (λ),M 1 (λ),M 2 (λ),...,M m−1 (λ) z + µ 0, 0, 0,...,M m (λ) z =0
(A.4)
while the structure of z is captured by:
⎛
⎞ ⎛
0 I 0
⎜
⎝
. .. . ..
⎟ ⎜
⎠ z = µ ⎝
0 0 I
I 0 0
. .. . ..
0 I 0
⎞
⎟
⎠ z.
(A.5)
Hence, problem (A.1) becomes equivalent to:
⎛⎛
⎜⎜
⎝⎝
⎛
µ
⎜
⎝
0 I 0
. .. . ..
0 0 I
M 0 (λ) M 1 (λ) ... M m−1 (λ)
⎞
⎟
⎠ +
⎞⎞
⎛
−I 0 0
0
. .. . ..
⎟⎟
0 −I 0 ⎠⎠ z = .
⎜
⎝ 0
0 ... 0 M m (λ)
0
⎞
⎟
⎠ .
(A.6)
Denoting the involved matrices in the latter equations by the matrices K(λ) and
L(λ), yields:
(
)
K(λ)+µL(λ) z =0.
(A.7)
This eigenvalue problem is linear in µ but still polynomial in λ. The dimensions of the
matrices K(λ) and L(λ) is(mn) × (m n). In a second step this problem is linearized
with respect to λ.
Note that Equation (A.6) can also be rewritten as follows:
⎛
⎞ ⎛
−µI I 0
. .. . ..
⎜
⎟
⎝ 0 −µI I ⎠ z = ⎜
⎝
M 0 (λ) M 1 (λ) ... M m−1 (λ)+µM m (λ)
0
.
.
0
0
⎞
⎟
⎠ .
(A.8)
A.2 Linearization with respect to λ
(
)
Let us expand the problem K(λ)+µL(λ) z = 0 in (A.7), which is linear in µ and
polynomial in λ, by writing the involved matrix as a polynomial in λ: