Linear Algebra, Theory And Applications, 2012a

410 POSITIVE MATRICES
( ) 2 ( ) 3 ( ) 4
λ
Now recall that r ∈ N was arbitrary and so this has shown that
λ 0
,
λ
λ 0
,
λ
λ 0
, ···
( )
A
are each eigenvalues of
λ 0
which has only finitely many and hence this sequence must
( )
λ
repeat. Therefore,
λ 0
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.
Then in this case, there exists a permutation matrix P such that
⎛ ⎞
v 1
. ( )
P v =
v r
u 0
≡ ≡ v 0 1
⎜ ⎟
⎝
. ⎠
0
Then
Therefore,
λ 0 v = A T v = A T P v 1 .
λ 0 v 1 = PA T P v 1 = Gv 1
Now P 2 = I because it is a permutation matrix. Therefore, the matrix G ≡ PA T P and A
are similar. Consequently, they have the same eigenvalues and it suffices from now on to
consider the matrix G rather than A. Then
( ) ( )( )
u M1 M
λ 0 =
2 u
0 M 3 M 4 0
where M 1 is r × r and M 4 is (n − r) × (n − r) . It follows from block multiplication and the
assumption that A and hence G are > 0that
( )
A
′
B
G =
.
0 C
Now let λ be an eigenvalue of G such that |λ| = λ 0 . Then from Lemma A.0.8, either
λ ∈ σ (A ′ )orλ ∈ σ (C) . Suppose without loss of generality that λ ∈ σ (A ′ ) . Since A ′ > 0
it has a largest positive eigenvalue λ ′ 0 which is obtained from (1.7). Thus λ ′ 0 ≤ λ 0 but λ
being an eigenvalue of A ′ , has its absolute value bounded by λ ′ 0 and so λ 0 = |λ| ≤λ ′ 0 ≤ λ 0
showing that λ 0 ∈ σ (A ′ ) . Now if there exists v >> 0 such that A ′T v = λ 0 v, then the first
part of this proof applies to the matrix A and so (λ/λ 0 ) is a root of unity. If such a vector,
v does not exist, then let A ′ playtheroleofA in the above argument and reduce to the
consideration of
( )
G ′ A
′′
B
≡
′
0 C ′
where G ′ is similar to A ′ and λ, λ 0 ∈ σ (A ′′ ) . Stop if A ′′T v = λ 0 v for some v >> 0.
Otherwise, decompose A ′′ similar to the above and add another prime. Continuing this way
you must eventually obtain the situation where (A ′···′ ) T v = λ 0 v for some v >> 0. Indeed,
this happens no later than when A ′···′ is a 1 × 1 matrix.