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10.3. THE KRONECKER CANONICAL FORM OF A MATRIX PENCIL 173
Working out the relationships for the coefficients of the powers of ρ 1 in (10.29),
yields the following in matrix-vector format:
⎛
⎞
B 0 ... ... 0
⎛
C B
.
0 C . .
. . ..
. 0 0 .. . ..
. ..
⎜
⎝
⎜ . . .
⎝ . . .. ⎟ B ⎠
0 0 ... ... C
z 0
−z 1
z 2
.
(−1) ε z ε
⎞ ⎛
= M ε ⎟ ⎜
⎠ ⎝
z 0
−z 1
z 2
.
(−1) ε z ε
⎞
=0, (10.30)
⎟
⎠
where the block matrix M ε has dimensions (ε +2)m × (ε +1)n and has rank <
(ε +1)n. Note that any non-zero vector in the kernel of the matrix M ε in (10.30)
yields a corresponding solution z(ρ 1 ) of degree ε in (10.27). Note furthermore that
the matrices M ε only depend on the known matrices B and C.
Now let the (algebraic) rank of a matrix M i be denoted by r i = rank(M i ). Because
ε was chosen as the least possible degree of z(ρ 1 ) for which a solution exists, it holds
for the matrices:
( B
M 0 =
C
⎛
⎞
B 0 ... 0
⎛ ⎞
)
B 0
C B
.
,M 1 = ⎝ C B ⎠ ,...,M ε−1 =
0 C . . . . ..
, (10.31)
0 C
⎜
⎝
.
.
. .. ⎟ B ⎠
0 0 ... C
that r 0 = n, r 1 =2n, ..., r ε−1 = εn.
Theorem 4 in Chapter 12 of [42] shows that if (10.27) has a solution z(ρ 1 )of
minimal degree ε and ε>0, then (B + ρ 1 C) is strictly equivalent to:
(
Lε 0
0 D + ρ 1 E
)
, (10.32)
where the submatrix L ε of dimension ε×(ε+1) exhibits the structure given in (10.18).
Furthermore, it holds that (D + ρ 1 E)z(ρ 1 ) = 0 has no solution z(ρ 1 ) of degree less
then ε.
The eigenvalue problem (B + ρ 1 C)ṽ = 0 is transformed using transformation
matrices V and W as follows:
W −1 (B + ρ 1 C) V (V −1 ṽ)=
(
Lε 0
0 D + ρ 1 E
) ( 0
w =
0
)
. (10.33)