13.6. CAUCHY’S INTERLACING THEOREM, EIGENVALUES 349

where U is the above unitary matrix. Thus the eigenvalues of Bk, µ̂1 < · · · < µ̂n−1 arestrictly increasing and µ̂ j ≡ µ j + 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(λ I−Ak) along the top row, the characteristic polynomial for Ak is then

q(λ ) = (λ −a)n−1

∏i=1

(λ − µ̂ i)−n−1

∑i=2|zi|2 (λ − µ̂1) · · · ̂(λ − µ̂ i) · · ·

(λ − µ̂n−1

)(13.12)

where ̂(λ − µ̂ i) indicates that this factor is omitted from the product ∏n−1i=1 (λ − µ̂ i) . To see

why this is so, consider the case where Bk is 3×3. In this case, you would have

(1 0T

0 U∗

)(λ I−Ak)

(1 0T

0 U

)=

λ −a z1 z2 z3

z1 λ − µ̂1 0 0z2 0 λ − µ̂2 0z3 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 − µ̂ i)− z1 det

 z1 0 0z2 λ − µ̂2 0z3 0 λ − µ̂3

+z2 det

 z1 λ − µ̂1 0z2 0 0z3 0 λ − µ̂3

− z3 det

 z1 λ − µ̂1 0z2 0 λ − µ̂2

z3 0 0



= (λ −a)3

∏i=1

(λ − µ̂ i)−

(|z1|2 (λ − µ̂2)(λ − µ̂3)+ |z2|2 (λ − µ̂1)(λ − µ̂3)

+ |z3|2 (λ − µ̂1)(λ − µ̂2)

)