Linear Algebra, Theory And Applications, 2012a

408 POSITIVE MATRICES
Lemma A.0.8 Let M be a matrix of the form
( )
A 0
M =
B C
or
M =
(
A B
0 C
where A is an r × r matrix and C is an (n − r) × (n − r) matrix. Then det (M) =
det (A)det(B) and σ (M) =σ (A) ∪ σ (C) .
)
Proof: To verify the claim about the determinants, note
( ) ( )(
A 0 A 0 I 0
=
B C 0 I B C
)
Therefore,
( A 0
det
B C
)
( A 0
=det
0 I
But it is clear from the method of Laplace expansion that
( ) A 0
det
=detA
0 I
) ( I 0
det
B C
and from the multilinear properties of the determinant and row operations that
( ) ( )
I 0
I 0
det
=det
=detC.
B C
0 C
ThecasewhereM is upper block triangular is similar.
This immediately implies σ (M) =σ (A) ∪ σ (C) .
Theorem A.0.9 Let A>0 and let λ 0 be given in (1.7). If λ is an eigenvalue for A such
that |λ| = λ 0 , then λ/λ 0 is a root of unity. Thus (λ/λ 0 ) m =1for some m ∈ N.
Proof: Applying Theorem A.0.7 to A T , there exists v > 0 such that A T v = λ 0 v. In
the first part of the argument it is assumed v >> 0. Now suppose Ax = λx, x ≠ 0 and that
|λ| = λ 0 . Then
A |x| ≥|λ||x| = λ 0 |x|
and it follows that if A |x| > |λ||x| , then since v >> 0,
λ 0 (v, |x|) < (v,A |x|) = ( A T v, |x| ) = λ 0 (v, |x|) ,
)
.
a contradiction. Therefore,
It follows that
∣ ∣∣∣∣∣
∑
A |x| = λ 0 |x| . (1.8)
and so the complex numbers,
j
A ij x j
∣ ∣∣∣∣∣
= λ 0 |x i | = ∑ j
A ij x j ,A ik x k
A ij |x j |