link to my thesis

10.3. THE KRONECKER CANONICAL FORM OF A MATRIX PENCIL 175
W as follows:
⎛
W =
⎜
⎝
Bz 1 ... Bz ε p ε+1 ... p m ,
⎟
⎠
⎞
⎛
V =
⎜
⎝
z 0 ... z ε q ε+1 ... q n−1 ⎟
⎠
⎞
(10.36)
where p ε+1 ,...,p m and q ε+1 ,...,q n−1 are basis column vectors to complete the matrices
up to square and invertible matrices.
Then it can easily be shown by using the relationships in (10.30) (i.e., Bz 0 =0,
Bz i+1 = Cz i , for i =0,...,ε−1, and (−1) ε Cz ε = 0), that the pencil W −1 (B+ρ 1 C)V
is equivalent with:
⎛
⎞
⎞
0 1 0 ... 0
1 0 ... 0 0
⎜
⎟ ⎜
⎟
0 0 1 ... 0 ∗
.
.
. ..
0 0 0 ... 1
0 0 0 ... 0
⎜
⎝ . . . . ∗
0 0 0 ... 0
⎛
0 1 ... 0 0 ∗
.
.
. ..
. ..
+ ρ 1 0 0 ... 1 0
0 0 ... 0 0
⎟ ⎜
⎠ ⎝ . . . . ∗
0 0 ... 0 0
. (10.37)
⎟
⎠
Now we conclude that this pencil has its upper left block equal to L ε and its lower
left block equal to 0, similar as in Equation (10.32).
Remark 10.1. The upper right block of the transformed matrix pencil (10.37) does
not equal a zero block as in (10.32) at this point. To make also this block zero, one
more transformation is required. We do not address this transformation here and
assume the upper right block of the pencil (10.37) to be zero without performing
this transformation. This is possible because the quasi diagonal blocks in Equation
(10.32) are not affected by this transformation. Moreover, when we continue splitting
off singular parts, we split off the upper right (non-zero) block, the bottom left (zero)
block and the structured block L ε to get rid of the unwanted eigenvalue(s). For details
see the third part of the proof of Theorem 4 in [42]. In fact, this amounts to choosing
p ε+1 ,...,p m and q ε+1 ,...,q n−1 in a special way.