176 CHAPTER 6. SPECTRAL THEORY
Proof: Denote by A (t) the matrix(atij)where if i ̸= j, atij = taij and atii = aii. Thus to
get A (t) multiply all non diagonal terms by t. Let t ∈ [0, 1] . Then A (0) = diag (a11, · · · , ann)and A (1) = A. Furthermore, the map, t → A (t) is continuous. Denote by Dt
j the Ger-
schgorin disc obtained from the jth row for the matrix A (t). Then it is clear that Dtj ⊆ Dj
the jth Gerschgorin disc for A. It follows aii is the eigenvalue for A (0) which is containedin the disc, consisting of the single point aii which is contained in Di. Letting K be theconnected component in Σ for Σ defined in Theorem 6.9.4 which is determined by aii, Ger-schgorin’s theorem implies that K ∩ σ (A (t)) ⊆ ∪n
j=1Dtj ⊆ ∪n
j=1Dj = Di ∪ (∪j ̸=iDj) andalso, since K is connected, there are not points of K in both Di and (∪j ̸=iDj) . Since at leastone point of K is in Di,(aii), it follows all of K must be contained in Di. Now by Theorem6.9.4 this shows there are points of K ∩ σ (A) in Di. The last assertion follows immediately.■
This can be improved even more. This involves the following lemma.
Lemma 6.9.6 In the situation of Theorem 6.9.4 suppose λ (0) = K0 ∩ σ (A (0)) and thatλ (0) is a simple root of the characteristic equation of A (0). Then for all t ∈ [0, 1] ,
σ (A (t)) ∩K0 = λ (t)
where λ (t) is a simple root of the characteristic equation of A (t) .
Proof: Let
S ≡ {t ∈ [0, 1] : K0 ∩ σ (A (s)) = λ (s) , a simple eigenvalue for all s ∈ [0, t]} .
Then 0 ∈ S so it is nonempty. Let t0 = sup (S) and suppose λ1 ̸= λ2 are two elements ofσ (A (t0))∩K0. Then choosing η > 0 small enough, and lettingDi be disjoint discs containingλi respectively, similar arguments to those of Lemma 6.9.3 can be used to conclude
Hi ≡ ∪s∈[t0−η,t0]σ (A (s)) ∩Di
is a connected and nonempty set for i = 1, 2 which would require that Hi ⊆ K0. Butthen there would be two different eigenvalues of A (s) contained in K0, contrary to thedefinition of t0. Therefore, there is at most one eigenvalue λ (t0) ∈ K0 ∩ σ (A (t0)) . Couldit be a repeated root of the characteristic equation? Suppose λ (t0) is a repeated root ofthe characteristic equation. As before, choose a small disc, D centered at λ (t0) and η smallenough that
H ≡ ∪s∈[t0−η,t0]σ (A (s)) ∩Dis a nonempty connected set containing either multiple eigenvalues of A (s) or else a singlerepeated root to the characteristic equation of A (s) . But since H is connected and containsλ (t0) it must be contained in K0 which contradicts the condition for s ∈ S for all theses ∈ [t0 − η, t0] . Therefore, t0 ∈ S as hoped. If t0 < 1, there exists a small disc centeredat λ (t0) and η > 0 such that for all s ∈ [t0, t0 + η] , A (s) has only simple eigenvalues inD and the only eigenvalues of A (s) which could be in K0 are in D. (This last assertionfollows from noting that λ (t0) is the only eigenvalue of A (t0) in K0 and so the others areat a positive distance from K0. For s close enough to t0, the eigenvalues of A (s) are eitherclose to these eigenvalues of A (t0) at a positive distance from K0 or they are close to theeigenvalue λ (t0) in which case it can be assumed they are in D.) But this shows that t0 isnot really an upper bound to S. Therefore, t0 = 1 and the lemma is proved. ■
With this lemma, the conclusion of the above corollary can be sharpened.
Corollary 6.9.7 Suppose one of the Gerschgorin discs, Di is disjoint from the union ofthe others. Then Di contains exactly one eigenvalue of A and this eigenvalue is a simpleroot to the characteristic polynomial of A.