Linear Algebra, Theory And Applications, 2012a

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|μ| ≤λ 0 . But also, the fact the entries of x all have the same sign shows μ = |μ| and so
μ ∈ S 1 .Sinceμ ≠ λ 0 , it must be that μ = |μ| 0
but it is not the case that x 0 +t Re y >> 0. Then A (x 0 + t Re y) >> 0 by Lemma A.0.2. But
this implies λ 0 (x 0 + t Re y) >> 0 which is a contradiction. Hence there exist real numbers,
α 1 and α 2 such that Re y = α 1 x 0 and Im y = α 2 x 0 showing that y =(α 1 + iα 2 ) x 0 . This
proves 3.
It is possible to obtain a simple corollary to the above theorem.
Corollary A.0.5 If A>0 and A m >> 0 for some m ∈ N, then all the conclusions of the
above theorem hold.
Proof: There exists μ 0 > 0 such that A m y 0 = μ 0 y 0 for y 0 >> 0 by Theorem A.0.4 and
μ 0 =sup{μ : A m x ≥ μx for some x ∈ K} .
Let λ m 0
= μ 0 . Then
(A − λ 0 I) ( A m−1 + λ 0 A m−2 + ···+ λ m−1
0 I ) y 0 =(A m − λ m 0 I) y 0 = 0
and so letting x 0 ≡ ( A m−1 + λ 0 A m−2 + ···+ λ m−1
0 I ) y 0 , it follows x 0 >> 0andAx 0 =
λ 0 x 0 .
Suppose now that Ax = μx for x ≠ 0 and μ ≠ λ 0 . Suppose |μ| ≥λ 0 . Multiplying both
sides by A, it follows A m x = μ m x and |μ m | = |μ| m ≥ λ m 0 = μ 0 and so from Theorem A.0.4,
since |μ m |≥μ 0 , and μ m is an eigenvalue of A m , it follows that μ m = μ 0 . But by Theorem
A.0.4 again, this implies x = cy 0 for some scalar, c and hence Ay 0 = μy 0 . Since y 0 >> 0,
it follows μ ≥ 0andsoμ = λ 0 , a contradiction. Therefore, |μ| 0 and A m >> 0 for some
∣∣( m ∈
)
N. Then for λ 0 given in (1.1),
∣∣ ∣∣
m ∣∣ A ∣∣ ∣∣
there exists a rank one matrix P such that lim m→∞ λ 0
− P =0.