186 CHAPTER 6. SPECTRAL THEORY

Now let {εk} be a decreasing sequence of very small positive numbers converging to 0 andlet Bk be defined by

U∗BkU = Dk, Dk ≡

µ1 + εk 0

µ2 + 2εk. . .

0 µn−1 + (n− 1) εk

where U is the above unitary matrix. Thus the eigenvalues of Bk, µ̂1 < · · · < µ̂n−1 arestrictly increasing and µ̂j ≡ µj + jεk. Let Ak be given by

Ak =

(a y∗

y Bk

)

Then(1 0∗

0 U∗

)Ak

(1 0∗

0 U

)=

(1 0∗

0 U∗

)(a y∗

y Bk

)(1 0∗

0 U

)

=

(a y∗

U∗y U∗Bk

)(1 0∗

0 U

)=

(a y∗U

U∗y Dk

)

We can replace y in the statement of the theorem with yk such that limk→∞ yk = y butzk ≡ U∗yk has the property that each component of zk is nonzero. This will probablytake place automatically but if not, make the change. This makes a change in Ak but stilllimk→∞Ak = A. The main part of this argument which follows has to do with fixed k.

Expanding det (λI −Ak) along the top row, the characteristic polynomial for Ak is then

q (λ) = (λ− a)

n−1∏i=1

(λ− µ̂i)−n−1∑i=2

|zi|2 (λ− µ̂1) · · · ̂(λ− µ̂i) · · ·(λ− µ̂n−1

)(6.27)

where ̂(λ− µ̂i) indicates that this factor is omitted from the product∏n−1

i=1 (λ− µ̂i) . To seewhy this is so, consider the case where Bk is 3× 3. In this case, you would have

(1 0T

0 U∗

)(λI −Ak)

(1 0T

0 U

)=

λ− a z1 z2 z3

z1 λ− µ̂1 0 0

z2 0 λ− µ̂2 0

z3 0 0 λ− µ̂3

In general, you would have an n× n matrix on the right with the same appearance. Thenexpanding as indicated, the determinant is

(λ− a)

3∏i=1

(λ− µ̂i)− z1 det

 z1 0 0

z2 λ− µ̂2 0

z3 0 λ− µ̂3

+z2 det

 z1 λ− µ̂1 0

z2 0 0

z3 0 λ− µ̂3

− z3 det

 z1 λ− µ̂1 0

z2 0 λ− µ̂2

z3 0 0

