392 APPENDIX B. POSITIVE MATRICES
Thanks to 2.2,
A
λ0P = x0v
T = P, P
(A
λ0
)= x0v
T
(A
λ0
)= x0v
T = P, (2.3)
andP 2 = x0v
Tx0vT = vTx0 = P. (2.4)
Therefore, (A
λ0− P
)2
=
(A
λ0
)2
− 2
(A
λ0
)P + P 2
=
(A
λ0
)2
− P.
Continuing this way, using 2.3 repeatedly, it follows((A
λ0
)− P
)m
=
(A
λ0
)m
− P. (2.5)
The eigenvalues of(
Aλ0
)− P are of interest because it is powers of this matrix which
determine the convergence of(
Aλ0
)mto P. Therefore, let µ be a nonzero eigenvalue of this
matrix. Thus ((A
λ0
)− P
)x = µx (2.6)
for x ̸= 0, and µ ̸= 0. Applying P to both sides and using the second formula of 2.3 yields
0 = (P − P )x =
(P
(A
λ0
)− P 2
)x = µPx.
But since Px = 0, it follows from 2.6 that
Ax = λ0µx
which implies λ0µ is an eigenvalue of A. Therefore, by Corollary B.0.5 it follows that eitherλ0µ = λ0 in which case µ = 1, or λ0 |µ| < λ0 which implies |µ| < 1. But if µ = 1, then x isa multiple of x0 and 2.6 would yield((
A
λ0
)− P
)x0 = x0
which says x0−x0vTx0 = x0 and so by 2.2, x0 = 0 contrary to the property that x0 >> 0.
Therefore, |µ| < 1 and so this has shown that the absolute values of all eigenvalues of(Aλ0
)− P are less than 1. By Gelfand’s theorem, Theorem 13.3.3, it follows
∣∣∣∣∣∣∣∣(( Aλ0)− P
)m∣∣∣∣∣∣∣∣1/m < r < 1
whenever m is large enough. Now by 2.5 this yields∣∣∣∣∣∣∣∣( Aλ0)m
− P
∣∣∣∣∣∣∣∣ = ∣∣∣∣∣∣∣∣(( Aλ0)− P
)m∣∣∣∣∣∣∣∣ ≤ rm