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174 CHAPTER 10. H 2 MODEL-ORDER REDUCTION FROM ORDER N TO N-2
By partitioning the new eigenvector w as w =
equations:
(
w1
w 2
)
, we get the following two
{
Lε w 1 =0
(D + ρ 1 E)w 2 =0
(10.34)
Note that also holds: ṽ = V
(
w1
w 2
)
. The first equation in (10.34) yields:
⎧
⎪⎨
⎪⎩
ρ 1 w 1,1 + w 1,2 =0
ρ 1 w 1,2 + w 1,3 =0
.
ρ 1 w 1,ε + w 1,ε+1 =0
(10.35)
When the first element w 1,1 of the vector w 1 is chosen equal to one, then the only
solution of (10.35) has the structure (1, −ρ 1 ,ρ 2 1,...,(−ρ 1 ) ε ) T . Therefore, any choice
of ρ 1 provides a non-trivial eigenvector
(
w 1
)
depending on ρ 1 : it generates an eigenvector
z for B + ρ 1 C through z = V , by choosing w 2 = 0, or by choosing a
w1
w 2
solution pair (ρ 1 ,w 2 ) to the second equation (D + ρ 1 E)w 2 = 0 and fixing ρ 1 to the
same value to generate w 1 . Therefore, this eigenvalue is recorded as an indeterminate
eigenvalue with multiplicity ε.
In [42] it is shown how to obtain the quasi block diagonal form (10.32) by choosing
appropriate transformation matrices W and V . Consider the pencil B + ρ 1 C as a
family of operators mapping C n to C m . With a suitable choice of bases the matrix
B + ρ 1 C takes up the form of (10.32). Such a suitable choice corresponds to taking
the vectors z 0 ,...,z ε , involved in a solution z(ρ 1 ) in Equation (10.27), as part of the
basis for C n . Subsequently, the vectors Bz 1 ,...,Bz ε are taken as part of the basis for
C m . This is allowed since {z 0 ,...,z ε } and {Bz 1 ,...,Bz ε } are linearly independent
sets (see [42]). Choosing the vectors z 0 ,...,z ε and Bz 1 ,...,Bz ε as the first columns
for the new bases in C n and C m , respectively, yields transformation matrices V and