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11.3. KRONECKER CANONICAL FORM OF A TWO-PARAMETER PENCIL 205
matrix-vector form as follows:
⎛
M η1,η 2 ⎜
⎝
z 0,0
.
z 0,η2
.
z η1,0
.
z η1,η 2
⎞
= 0 (11.27)
⎟
⎠
where the matrix M η1,η 2
is a block matrix containing the coefficient matrices B T ,
C T , and D T from Equation (11.7). The matrix M η1,η 2
is independent of ρ 1 and ρ 2
and has a dimension of (η 1 +2)(η 2 +2)m × (η 1 +1)(η 2 +1)n.
Any non-zero vector in the kernel of the matrix M η1,η 2
in (11.27) yields a corresponding
solution z(ρ 1 ,ρ 2 ) of degree at most η 1 and η 2 for ρ 1 and ρ 2 , respectively.
Therefore, a computation of the kernel of the matrix M η1,η 2
suffices to compute a
vector z(ρ 1 ,ρ 2 ) in (11.22).
Example 11.1. To illustrate the technique described above, a small example is given.
Note that this example only serves the purpose of making clear how to construct the
block matrix M η1,η 2
in (11.27).
Suppose the two-parameter linear matrix pencil B T +ρ 1 C T +ρ 2 D T (with matrices
of dimensions m × n) is given and one wants to check whether there is a solution
z(ρ 1 ,ρ 2 )of(B T + ρ 1 C T + ρ 2 D T )z(ρ 1 ,ρ 2 ) = 0 where the degree of ρ 1 is η 1 = 1 and
the degree of ρ 2 is η 2 =2.
First, the expression (B T + ρ 1 C T + ρ 2 D T )z(ρ 1 ,ρ 2 ) = 0 is written as (P (ρ 2 ) T +
ρ 1 Q(ρ 2 ) T ) z(ρ 1 ,ρ 2 ) = 0 where P (ρ 2 ) T = B T + ρ 2 D T and Q(ρ 2 ) T = C T .
We know that every solution z(ρ 1 ,ρ 2 ) should take the form z 0 (ρ 2 ) − ρ 1 z 1 (ρ 2 ),
because η 1 = 1. If such a solution is substituted into (P (ρ 2 ) T +ρ 1 Q(ρ 2 ) T ) z(ρ 1 ,ρ 2 )=
0, and the relationships for the coefficients of the powers of ρ 1 are worked out, it can
be written as:
⎛
P (ρ 2 ) T ⎞
0 ( )
⎝ Q(ρ 2 ) T P (ρ 2 ) T ⎠ z0 (ρ 2 )
0 Q(ρ 2 ) T −z 1 (ρ 2 )
(
z0 (ρ
= M 1,2 (ρ 1 ,ρ 2 )
2 )
−z 1 (ρ 2 )
)
= 0 (11.28)
where the dimension of the block matrix M 1,2 (ρ 1 ,ρ 2 )is(η 1 +2) m × (η 1 +1) n =
3 m × 2 n.
The degree of ρ 2 is chosen here as 2 and therefore z i (ρ 2 ) can be written as
z i (ρ 2 )=z i,0 + ρ 2 z i,1 + ρ 2 2z i,2 for i =0, 1. This can be substituted into (11.28).
When the relationships for the coefficients are worked out and P (ρ 2 ) T = B T + ρ 2 D T