Now let A be an n×n matrix. Recall that the eigenvalues of A are given by the zeros of the polynomial,
p_{A}
(z)
= det
(zI − A)
where I is the n × n identity. You can argue that small changes in A will produce
small changes in p_{A}
(z)
and p_{A}^{′}
(z)
. Let γ_{k} denote a very small closed circle which winds around z_{k}, one of
the eigenvalues of A, in the counter clockwise direction so that n
(γk,zk)
= 1. This circle is to enclose only
z_{k} and is to have no other eigenvalue on it. Then apply Theorem 14.4.1. According to this
theorem
1 ∫ p′A(z)
2πi p-(z)dz
γ A
is always an integer equal to the multiplicity of z_{k} as a root of p_{A}
(t)
. Therefore, small changes in A result
in no change to the above contour integral because it must be an integer and small changes in A result in
small changes in the integral. Therefore whenever B is close enough to A, the two matrices have the same
number of zeros inside γ_{k}, the zeros being counted according to multiplicity. By making the radius of the
small circle equal to ε where ε is less than the minimum distance between any two distinct eigenvalues of
A, this shows that if B is close enough to A, every eigenvalue of B is closer than ε to some eigenvalue of A.
■
Theorem 14.4.2If λ is an eigenvalue of A, then ifall the entries of B are close enough to thecorresponding 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 for
t ∈
[0,1]
.
Lemma 14.4.3Let λ
(t)
∈ σ
(A (t))
for t < 1 and let Σ_{t} = ∪_{s≥t}σ
(A(s))
. Also let K_{t}be theconnected component of λ
(t)
in Σ_{t}. Then there exists η > 0 such that K_{t}∩ σ
(A (s))
≠∅ for alls ∈
[t,t+ η]
.
Proof: Denote by D
(λ(t),δ)
the disc centered at λ
(t)
having radius δ > 0, with other occurrences 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 in D
(λ(t),δ)
and that
p_{A(t)
} has no zeroes on the boundary of this disc. Then by continuity, and the above discussion and
theorem, there exists η > 0,t + η < 1, such that for s ∈
[t,t+ η]
, p_{A(s)
} also has no zeroes on the
boundary of this disc and A
(s)
has the same number of 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 = σ(A (s))∩ D (λ(t),δ) .
s∈[t,t+η]
It will be shown that H is connected. Suppose not. Then H = P ∪Q where P,Q are separated and λ
(t)
∈ P.
Let s_{0}≡ inf
{s : λ (s) ∈ Q for some λ (s) ∈ σ (A(s))}
. There exists λ
(s0)
∈ σ
(A (s0))
∩ D
(λ (t),δ)
. If
λ
(s0)
∕∈
Q, then from the above discussion there are λ
(s)
∈ σ
(A (s))
∩ Q for s > s_{0} arbitrarily close to
λ
(s0)
. Therefore, λ
(s0)
∈ Q which shows that s_{0}> t because λ
(t)
is the only element of σ
(A (t))
in
D
(λ(t),δ)
and λ
(t)
∈ P. Now let s_{n}↑ s_{0}. Then λ
(sn)
∈ P for any λ
(sn)
∈ σ
(A (sn))
∩ D
(λ(t),δ)
and also it follows from the above discussion that for some choice of s_{n}→ s_{0}, λ
(sn)
→ λ
(s0)
which contradicts P and Q separated and nonempty. Since P is nonempty, this shows Q = ∅.
Therefore, H is connected as claimed. But K_{t}⊇ H and so K_{t}∩ σ
(A(s))
≠∅ for all s ∈
[t,t+ η]
.■
Theorem 14.4.4Suppose A
(t)
is an n×n matrix and that t → A
(t)
is continuous for t ∈
[0,1]
.Let λ
(0)
∈ σ
(A(0))
and define Σ ≡ ∪_{t∈[0,1]
}σ
(A(t))
. Let K_{λ(0)
} = K_{0}denote the connectedcomponent of λ
(0)
in Σ. Then K_{0}∩ 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 t_{0} = sup
(S)
. Say
σ
(A (t0))
= λ_{1}
(t0)
,
⋅⋅⋅
,λ_{r}
(t0)
.
Claim: At least one of these is a limit point of K_{0} and consequently must be in K_{0} which
shows that S has a last point. Why is this claim true? Let s_{n}↑ t_{0} so s_{n}∈ S. Now let the discs,
D
(λi(t0),δ)
,i = 1,
⋅⋅⋅
,r be disjoint with p_{A(t0)
} having no zeroes on γ_{i} the boundary of D
(λi(t0),δ)
.
Then for n large enough it follows from Theorem 14.4.1 and the discussion following it that σ
(A (sn))
is
contained in ∪_{i=1}^{r}D
(λi(t0),δ)
. It follows that K_{0}∩
(σ (A (t0))+ D (0,δ))
≠∅ for all δ small
enough. This requires at least one of the λ_{i}
(t0)
to be in K_{0}. Therefore, t_{0}∈ S and S has a last
point.
Now by Lemma 14.4.3, if t_{0}< 1, then K_{0}∪K_{t} would be a strictly larger connected set containing λ
(0)
.
(The reason this would be strictly larger is that K_{0}∩ σ
(A (s))
= ∅ for some s ∈
(t,t+ η)
while
K_{t}∩ σ
(A (s))
≠∅ for all s ∈
[t,t+ η]
.) Therefore, t_{0} = 1. ■
Corollary 14.4.5Suppose one of the Gerschgorin discs, D_{i}is disjoint from the union of the others.Then D_{i}contains an eigenvalue of A. Also, if there are n disjoint Gerschgorin discs, then each onecontains an eigenvalue of A.
Proof:Denote by A
(t)
the matrix
(at)
ij
where if i≠j, a_{ij}^{t} = ta_{ij} and a_{ii}^{t} = a_{ii}. 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 D_{j}^{t} the Gerschgorin disc obtained from the j^{th}
row for the matrix A
(t)
. Then it is clear that D_{j}^{t}⊆ D_{j} the j^{th} Gerschgorin disc for A. It follows a_{ii} is the
eigenvalue for A
(0)
which is contained in the disc, consisting of the single point a_{ii} which is contained in
D_{i}. Letting K be the connected component in Σ for Σ defined in Theorem 14.4.4 which is determined by
a_{ii}, Gerschgorin’s theorem implies that K ∩ σ
(A(t))
⊆∪_{j=1}^{n}D_{j}^{t}⊆∪_{j=1}^{n}D_{j} = D_{i}∪
(∪j⁄=iDj)
and
also, since K is connected, there are not points of K in both D_{i} and
(∪j⁄=iDj )
. Since at least
one point of K is in D_{i},(a_{ii}), it follows all of K must be contained in D_{i}. Now by Theorem
14.4.4 this shows there are points of K ∩ σ
(A )
in D_{i}. The last assertion follows immediately.
■
This can be improved even more. This involves the following lemma.
Lemma 14.4.6In the situation of Theorem 14.4.4suppose λ
(0)
= K_{0}∩ σ
(A(0))
and that λ
(0)
is asimple 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 t_{0} = sup
(S )
and suppose λ_{1}≠λ_{2} are two elements of σ
(A (t0))
∩K_{0}. Then choosing
η > 0 small enough, and letting D_{i} be disjoint discs containing λ_{i} respectively, similar arguments to those
of Lemma 14.4.3 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 H_{i}⊆ K_{0}. But then there would be
two different eigenvalues of A
(s)
contained in K_{0}, contrary to the definition of t_{0}. Therefore, there is at
most one eigenvalue λ
(t)
0
∈ K_{0}∩ σ
(A (t))
0
. Could it be a repeated root of the characteristic equation?
Suppose λ
(t)
0
is a repeated root of the characteristic equation. As before, choose a small disc, D centered
at λ
(t)
0
and η small enough that
H ≡ ∪s∈[t0−η,t0]σ (A(s))∩ D
is a nonempty connected set containing either multiple eigenvalues of A
(s)
or else a single repeated root to
the characteristic equation of A
(s)
. But since H is connected and contains λ
(t )
0
it must be contained in
K_{0} which contradicts the condition for s ∈ S for all these s ∈
[t − η,t ]
0 0
. Therefore, t_{0}∈ S as hoped. If
t_{0}< 1, there exists a small disc centered at λ
(t)
0
and η > 0 such that for all s ∈
[t,t + η]
0 0
, A
(s)
has
only simple eigenvalues in D and the only eigenvalues of A
(s)
which could be in K_{0} are in D.
(This last assertion follows from noting that λ
(t )
0
is the only eigenvalue of A
(t )
0
in K_{0} and so
the others are at a positive distance from K_{0}. For s close enough to t_{0}, the eigenvalues of
A
(s)
are either close to these eigenvalues of A
(t )
0
at a positive distance from K_{0} or they
are close to the eigenvalue λ
(t )
0
in which case it can be assumed they are in D.) But this
shows that t_{0} is not really an upper bound to S. Therefore, t_{0} = 1 and the lemma is proved.
■
With this lemma, the conclusion of the above corollary can be sharpened.
Corollary 14.4.7Suppose one of the Gerschgorin discs, D_{i}is disjoint from the union of the others.Then D_{i}contains exactly one eigenvalue of A and this eigenvalue is a simple root to the characteristicpolynomial of A.
Proof: In the proof of Corollary 14.4.5, note that a_{ii} is a simple root of A
(0)
since otherwise the i^{th}
Gerschgorin disc would not be disjoint from the others. Also, K, the connected component determined by
a_{ii} must be contained in D_{i} because it is connected and by Gerschgorin’s theorem above, K ∩ σ
(A (t))
must be contained in the union of the Gerschgorin discs. Since all the other eigenvalues of A
(0)
, the a_{jj},
are outside D_{i}, it follows that K ∩σ
is disjoint from the other
discs. Therefore, there should be an eigenvalue in D
(5,1)
. The actual eigenvalues are not easy to
find. They are the roots of the characteristic equation, t^{3}− 6t^{2} + 3t + 5 = 0. The numerical
values of these are −.66966,1.4231, and 5.24655, verifying the predictions of Gerschgorin’s
theorem.