Page 39 - Textos de Matemática Vol. 44

WIELANDT’S THEOREM AND A RESOLVENT ESTIMATE
S. W. DRURY
Dedicated to Nat´alia Bebiano on the occasion of her 60th birthday.
Abstract. Let L =] − ∞,−1] ∪ [1,∞[ viewed as a subset of C. Let z ∈ C and r ≥ 0. Let D be the closed disk, centred at z of radius r and assume that D ∩ L = ∅. Let ||| ||| denote the usual operator norm. We obtain an upper bound for |||(I − ZH)−1|||, where H is a hermitian matrix with its spectrum in [−1, 1] and Z is a matrix with |||Z − zI||| ≤ r.
1. Introduction
The motivation for this note is a celebrated theorem of Helmut Wielandt
[1].
Theorem 1.1. Let a1,a2,...,an and b1,b2,...,bn be given sets of complex numbers. Then the following are equivalent:
• There exist normal matrices A and B with the given numbers as eigen- values sets and A − B singular.
• It is impossible to separate {a1,a2,...,an} from {b1,b2,...,bn} in the complex plane C with a straight line or circle.
In particular, the theorem asserts that if A and B are normal matrices whose spectra can be separated in C with a straight line or circle, then A − B is invertible. Our objective here is, in a particular case, to quantify that result. Let L =] − ∞, −1] ∪ [1, ∞[ viewed as a subset of C. Let z ∈ C and r ≥ 0. Let D be the closed disk, centred at z of radius r and assume that D ∩ L = ∅. Then it is clear geometrically that D and L can be separated by a circle, in fact, a circle centred at z with radius slightly bigger than r. In this case, Wielandt’s Theorem asserts that if Z is a normal matrix with spectrum in D and if K is a hermitian matrix with spectrum in L, then K − Z is invertible. It then follows that I − ZK−1 is also invertible.
2010 Mathematics Subject Classification. 15A60, 47A10
Key words and phrases. matrix, hermitian, resolvent, operator norm
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