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70 Chapter 3. Numeric solutions
Proof. Let us first note that R(P ) > 0 implies the inequality PSP > PA + DP + Q.
We assume the Newton sequence to be initialized with K0 = 0 then the first element K1 is calculated via the Sylvester equation
−K1A−DK1 =Q,
which can be rewritten as
[(−AT)⊗I2n +In ⊗(−D)]K⃗1 =Q⃗,
where

= Λ0
⃗:Rk×l → Rkl×1
X = x1 ... xl → xT1 ,...,xTl T
and ⊗ denotes the Kronecker product. Since −A is an M-matrix the matrix Λ0 is also an M-matrix and hence by Theorem 3.1.1, (ii), we infer K1 ≥ 0. Now define for i ∈ N0
Λi :=[(−A+SKi)T ⊗I2n +In ⊗(−D+KiS)]. We prove by induction the following properties for i ∈ N0:
1. Ki ≤ Ki+1,
2. Ki ≤P,
3. Λi is M-matrix.
As basis of induction we use what we proved for i = 0 above. We assume that the properties 1.–3. hold for an arbitrarily given i ≥ 0. To prove 2. for i + 1 we use the definition of Newton’s iteration and the property R(P ) > 0 and write
(P − Ki+1)(A − SKi) + (D − KiS)(P − Ki+1)
= PA+DP −PSKi −KiSP +KiSKi +Q
≤ PSP − PSKi − KiSP + KiSKi
= (P−Ki)S(P−Ki)
≥0 ≥0 ≤ 0.
Since Λi is M-matrix we conclude immediately, that P − Ki+1 ≥ 0, hence P ≥ Ki+1.