6.9. ADVANCED THEOREMS 175

Theorem 6.9.2 If λ is an eigenvalue of A, then if all the entries of B are close enough tothe corresponding entries of A, some eigenvalue of B will be within ε of λ.

Consider the situation that A (t) is an n×n matrix and that t→ A (t) is continuous fort ∈ [0, 1] .

Lemma 6.9.3 Let λ (t) ∈ σ (A (t)) for t < 1 and let Σt = ∪s≥tσ (A (s)) . Also let Kt be theconnected component of λ (t) in Σt. Then there exists η > 0 such that Kt ∩σ (A (s)) ̸= ∅ forall s ∈ [t, t+ η] .

Proof: Denote by D (λ (t) , δ) the disc centered at λ (t) having radius δ > 0, with otheroccurrences of this notation being defined similarly. Thus

D (λ (t) , δ) ≡ {z ∈ C : |λ (t)− z| ≤ δ} .

Suppose δ > 0 is small enough that λ (t) is the only element of σ (A (t)) contained inD (λ (t) , δ) and that pA(t) has no zeroes on the boundary of this disc. Then by continuity, andthe above discussion and theorem, there exists η > 0, t+ η < 1, such that for s ∈ [t, t+ η] ,pA(s) also has no zeroes on the boundary of this disc and A (s) has the same numberof eigenvalues, counted according to multiplicity, in the disc as A (t) . Thus σ (A (s)) ∩D (λ (t) , δ) ̸= ∅ for all s ∈ [t, t+ η] . Now let

H =⋃

s∈[t,t+η]

σ (A (s)) ∩D (λ (t) , δ) .

It will be shown that H is connected. Suppose not. Then H = P ∪ Q where P,Q areseparated and λ (t) ∈ P. Let s0 ≡ inf {s : λ (s) ∈ Q for some λ (s) ∈ σ (A (s))} . There existsλ (s0) ∈ σ (A (s0)) ∩ D (λ (t) , δ) . If λ (s0) /∈ Q, then from the above discussion there areλ (s) ∈ σ (A (s))∩Q for s > s0 arbitrarily close to λ (s0) . Therefore, λ (s0) ∈ Q which showsthat s0 > t because λ (t) is the only element of σ (A (t)) in D (λ (t) , δ) and λ (t) ∈ P. Nowlet sn ↑ s0. Then λ (sn) ∈ P for any λ (sn) ∈ σ (A (sn))∩D (λ (t) , δ) and also it follows fromthe above discussion that for some choice of sn → s0, λ (sn) → λ (s0) which contradicts Pand Q separated and nonempty. Since P is nonempty, this shows Q = ∅. Therefore, H isconnected as claimed. But Kt ⊇ H and so Kt ∩ σ (A (s)) ̸= ∅ for all s ∈ [t, t+ η] . ■

Theorem 6.9.4 Suppose A (t) is an n × n matrix and that t → A (t) is continuous fort ∈ [0, 1] . Let λ (0) ∈ σ (A (0)) and define Σ ≡ ∪t∈[0,1]σ (A (t)) . Let Kλ(0) = K0 denote theconnected component of λ (0) in Σ. Then K0 ∩ σ (A (t)) ̸= ∅ for all t ∈ [0, 1] .

Proof: Let S ≡ {t ∈ [0, 1] : K0 ∩ σ (A (s)) ̸= ∅ for all s ∈ [0, t]} . Then 0 ∈ S. Let t0 =sup (S) . Say σ (A (t0)) = λ1 (t0) , · · · , λr (t0) .

Claim: At least one of these is a limit point of K0 and consequently must be in K0

which shows that S has a last point. Why is this claim true? Let sn ↑ t0 so sn ∈ S.Now let the discs, D (λi (t0) , δ) , i = 1, · · · , r be disjoint with pA(t0) having no zeroes on γithe boundary of D (λi (t0) , δ) . Then for n large enough it follows from Theorem 6.9.1 andthe discussion following it that σ (A (sn)) is contained in ∪r

i=1D (λi (t0) , δ). It follows thatK0 ∩ (σ (A (t0)) +D (0, δ)) ̸= ∅ for all δ small enough. This requires at least one of theλi (t0) to be in K0. Therefore, t0 ∈ S and S has a last point.

Now by Lemma 6.9.3, if t0 < 1, then K0 ∪Kt would be a strictly larger connected setcontaining λ (0) . (The reason this would be strictly larger is that K0 ∩ σ (A (s)) = ∅ forsome s ∈ (t, t+ η) while Kt ∩ σ (A (s)) ̸= ∅ for all s ∈ [t, t+ η].) Therefore, t0 = 1. ■

Corollary 6.9.5 Suppose one of the Gerschgorin discs, Di is disjoint from the union ofthe others. Then Di contains an eigenvalue of A. Also, if there are n disjoint Gerschgorindiscs, then each one contains an eigenvalue of A.