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CHAPTER 2. PRELIMINARIES 32
Lemma 2.15. Assume that A is positive definite, C positive semi-definite, and that if B
does not possess full row-rank then C is positive definite. Then the generalized eigenvalues
λ of ( ( ) ( ( )
A B
T u A 0 u
= λ
B −C)
p 0 S)
p
are contained in the set [
Proof. To solve the problem
−1; 1 − √ 5
2
]
∪
[
1; 1 + √ 5
2
λAu = Au + B T p
λSp = Bu − Cp
]
.
(2.40a)
(2.40b)
we distinguish two cases.
First if λ = 1 then (2.40a) can be fulfilled with p = 0 and (2.40a) by any u with Bu = 0.
Now for λ ≠ 1 we get u = 1
λ−1 A−1 B T p from (2.40a), insert this in (2.40b) and get
(
λ − 1 + √ 5
2
) (
λ − 1 − √ 5
2
)
BA −1 B T p + (λ + 1)(λ − 1)Cp = 0. (2.41)
As both BA −1 B T and C are positive semi-definite and at least one of them is positive
definite, (2.41) can only be fulfilled for
[
λ ∈ −1; 1 − √ ] (
5
∪ 1; 1 + √ ]
5
,
2
2
which completes the proof.
Remark 2.16. If C ≡ 0 then one can even show that λ ∈ {1, (1 ± √ 5/2}. Similar and more
general results can be found e.g. in [RW92], [IRT93], or [SW94].
and
Now we can easily calculate that
and deduce from Lemma 2.15 that
and
‖ ˆK −1 K‖ ˆK = ‖K1/2 ˆK−1 K 1/2 ‖ l2
‖K −1 ˆK‖ ˆK = ‖ ˆK 1/2 K −1 ˆK1/2 ‖ l2 ,
‖K 1/2 ˆK−1 K 1/2 ‖ l2 ≤ 1 + √ 5
,
2
‖ ˆK 1/2 K −1 ˆK1/2 ‖ l2 ≤
2
√
5 − 1
.
Thus we can estimate
( )
cond ˆK−1 ˆK
K ≤ 3 + √ 5
,
2
where the upper bound is clearly independent of h.