395
Also, summing 2.9 on j and using that λ is an eigenvalue for x, it follows from 2.8 that
λxi =∑j
Aijxj = µi
∑j
Aij |xj | = µiλ0 |xi| . (2.11)
From 2.10 and 2.11, ∑j
Aijxjµrj = µi
∑j
Aij |xj |µrj
= µi
∑j
Aij
see 2.11︷ ︸︸ ︷µj |xj |µr−1
j
= µi
∑j
Aij
(λ
λ0
)xjµ
r−1j
= µi
(λ
λ0
)∑j
Aijxjµr−1j
Now from 2.10 with r replaced by r − 1, this equals
µ2i
(λ
λ0
)∑j
Aij |xj |µr−1j = µ2
i
(λ
λ0
)∑j
Aijµj |xj |µr−2j
= µ2i
(λ
λ0
)2∑j
Aijxjµr−2j .
Continuing this way, ∑j
Aijxjµrj = µk
i
(λ
λ0
)k∑j
Aijxjµr−kj
and eventually, this shows∑j
Aijxjµrj = µr
i
(λ
λ0
)r∑j
Aijxj
=
(λ
λ0
)r
λ (xiµri )
and this says(
λλ0
)r+1
is an eigenvalue for(
Aλ0
)with the eigenvector being
(x1µr1, · · · , xnµr
n)T.
Now recall that r ∈ N was arbitrary and so this has shown that(
λλ0
)2,(
λλ0
)3,(
λλ0
)4, · · ·
are each eigenvalues of(
Aλ0
)which has only finitely many and hence this sequence must
repeat. Therefore,(
λλ0
)is a root of unity as claimed. This proves the theorem in the case
that v >> 0.Now it is necessary to consider the case where v > 0 but it is not the case that v >> 0.