56 Dynamical Systems
(ii) In the case of the convergence of (A/λ)
t
, every column of the limit
matrix is a strictly positive column eigenvector of A and every row is a
strictly positive row eigenvector of A , both associated with λ.
Proof. Proposition 9.2 ensures that λ>0, so that ( A/λ) makes sense.
Moreover, the dominant root of ( A/λ) is unity. This enables us to assume,
without loss of generality, that λ(A) = 1.
First we shall prove (ii). To this end, let B = lim A
t
. Clearly, B ≥ 0.
We also have lim A
t+1
= B, so that AB = A lim A
t
= lim A
t+1
= B
and, similarly, BA = B. AB = B implies that Ab = b for any column
of B. It follows, by Remark 9.1, that b is either a strictly positive col-
umn eigenvector of A associated with λ = 1 or the column zero vector.
Likewise BA = B implies that any row of B is either a strictly positive
row eigenvector of A associated with λ = 1 or the row zero vector. In
fact, B = (b
ij
) is a strictly positive matrix. For if some b
ij
= 0, the above
results imply that both the i th row and the jth column of B vanish, and
hence, all the rows and columns vanish. But the possibility of B = 0is
ruled out, because Bx
0
= lim A
t
x
0
= x
0
for a strictly positive column
eigenvector x
0
of A associated with λ = 1. This proves (ii).
The “only if ” part of (i) immediately follows from (ii). In fact, if
lim A
t
= B exists, B must be strictly positive by (ii). Hence A
t
must also
be strictly positive for t ≥ k, whenever k is large enough. For proof of the
“if ” part of (i), first assume that the assertion is true for any strictly pos-
itive matrix. Then, since A
k
>> 0 and λ(A
k
) = λ( A)
k
= 1 by Remark
9.1, lim(A
k
)
t
= B >> 0 exists. Every positive integer t is uniquely ex-
pressible in the form t = p(t)k + r(t), where p(t), r(t) are nonnegative
integers with 0 ≤ r(t) < k, lim p(t) =+∞. Therefore, letting (A
k
)
p(t )
=
B + R(t), we have lim R(t) = 0. Hence A
t
= A
r(t )
(A
k
)
p(t )
= A
r(t )
(B +
R(t)) = A
r(t )
B + A
r(t )
R(t) → B because A
r(t )
B = B and A
r(t )
is
bounded. The reason for A
r(t )
B = B is that since A and A
k
have eigen-
vectors in common, the columns of B are strictly positive column eigen-
vectors of A associated with λ = 1; hence A
r(t )
B = AA···AB = B.
Thus it remains to prove the assertion for strictly positive matrices.
Now let A be a strictly positive matrix with λ(A) = 1. To prove the
convergence of A
t
, we only have to show that the vector sequence {A
t
y}
is convergent for any positive y > 0.
Set y
t
= A
t
y for any given y > 0. y
t
is a solution of the difference
equation
y
t+1
= Ay
t
, y
0
= y.