391

Assume the first holds. Then varying t ∈ R, there exists a value of t such that x0+tRey > 0but it is not the case that x0+tRey >> 0. Then A (x0 + tRey) >> 0 by Lemma B.0.2. Butthis implies λ0 (x0 + tRey) >> 0 which is a contradiction. Hence there exist real numbers,α1 and α2 such that Rey = α1x0 and Imy = α2x0 showing that y =(α1 + iα2)x0. Thisproves 3.

It is possible to obtain a simple corollary to the above theorem.

Corollary B.0.5 If A > 0 and Am >> 0 for some m ∈ N, then all the conclusions of theabove theorem hold.

Proof: There exists µ0 > 0 such that Amy0 = µ0y0 for y0 >> 0 by Theorem B.0.4 and

µ0 = sup {µ : Amx ≥ µx for some x ∈ K} .

Let λm0 = µ0. Then

(A− λ0I)(Am−1 + λ0A

m−2 + · · ·+ λm−10 I

)y0 = (Am − λm0 I)y0 = 0

and so letting x0 ≡(Am−1 + λ0A

m−2 + · · ·+ λm−10 I

)y0, it follows x0 >> 0 and Ax0 =

λ0x0.Suppose now that Ax = µx for x ̸= 0 and µ ̸= λ0. Suppose |µ| ≥ λ0. Multiplying both

sides by A, it follows Amx = µmx and |µm| = |µ|m ≥ λm0 = µ0 and so from Theorem B.0.4,since |µm| ≥ µ0, and µ

m is an eigenvalue of Am, it follows that µm = µ0. But by TheoremB.0.4 again, this implies x = cy0 for some scalar, c and hence Ay0 = µy0. Since y0 >> 0,it follows µ ≥ 0 and so µ = λ0, a contradiction. Therefore, |µ| < λ0.

Finally, if Ax = λ0x, then Amx = λm0 x and so x = cy0 for some scalar, c. Consequently,(

Am−1 + λ0Am−2 + · · ·+ λm−1

0 I)x = c

(Am−1 + λ0A

m−2 + · · ·+ λm−10 I

)y0

= cx0.

Hencemλm−1

0 x = cx0

which shows the dimension of the eigenspace for λ0 is one. ■The following corollary is an extremely interesting convergence result involving the pow-

ers of positive matrices.

Corollary B.0.6 Let A > 0 and Am >> 0 for some m ∈ N. Then for λ0 given in 2.1,

there exists a rank one matrix P such that limm→∞

∣∣∣∣∣∣( Aλ0

)m− P

∣∣∣∣∣∣ = 0.

Proof: Considering AT , and the fact that A and AT have the same eigenvalues, CorollaryB.0.5 implies the existence of a vector, v >> 0 such that

ATv = λ0v.

Also let x0 denote the vector such that Ax0 = λ0x0 with x0 >> 0. First note that xT0 v > 0

because both these vectors have all entries positive. Therefore, v may be scaled such that

vTx0 = xT0 v = 1. (2.2)

DefineP ≡ x0v

T .