The four row Gershgorin discs have centres

9, 3, - 4, - 5

and radii

1, 2, 2, 1

, respectively. Their real-axis intersections are

[8, 10], [1, 5], [- 6, - 2], [- 6, - 4] .

(1)

None of the Gershgorin discs contains zero, so the Gershgorin circle theorem gives

0 \in / spec (A), A is nonsingular .

(2)

The discs centred at

9

and

3

are each isolated. By Gershgorin eigenvalue counting, they contain exactly one eigenvalue apiece, and both eigenvalues are simple. Each must be real: the characteristic polynomial is real, so a nonreal root would bring a distinct conjugate root into the same disc. Denote these eigenvalues by

β \in [8, 10]

and

α \in [1, 5]

. The remaining two eigenvalues lie in the union of the negative-centred discs.

For the shift

4

, the distances satisfy

∣ α - 4∣ \leq 3, ∣ β - 4∣ \geq 4, ∣ γ - 4∣ \geq 6

(3)

for either of the other eigenvalues

γ

, since its real part is at most

- 2

. Thus

α

is uniquely closest to

4

. Once

4

is checked not to be an eigenvalue, the natural full target-disc interval enclosure and a conservative asymptotic direction factor are

[a, b] = [1, 5], q_{4} \leq \frac{3}{4} .

(4)

As always, inverse iteration requires a nonzero initial target component. The following calculation verifies the shift and gives sharper information.

For the tridiagonal matrix, let

p_{j} (t)

be the determinant of the leading

j \times j

block of

t I - A

. Expansion along the last row gives

p_{0} (t) p_{2} (t) p_{3} (t) p_{4} (t) = 1, = (t - 3) (t - 9) + 1 = t^{2} - 12 t + 28, = (t + 4) p_{2} (t) - p_{1} (t) = t^{3} - 8 t^{2} - 21 t + 121, = (t + 5) p_{3} (t) - p_{2} (t) = t^{4} - 3 t^{3} - 62 t^{2} + 28 t + 577. p_{1} (t) = t - 9,

(5)

In particular,

det A = 577

p_{4} (4) = - 239

and

p_{4} (5) = - 583

. Both shifted linear systems are nonsingular. Further exact evaluations are

t p_{4} (t) - 6 121 - 5 - 113 - 4 - 79 - 3 97 3103 4 - 239 8 - 607 9 181.

(6)

The intermediate value theorem gives a real root in each of

(- 6, - 5)

(- 4, - 3)

(3, 4)

and

(8, 9)

. As

p_{4}

has polynomial degree four, these are all its roots and each is simple. In particular,

A

is a diagonalizable matrix,

α \in (3, 4)

and

β \in (8, 9)

. The closest competitor to

α

at the shift

4

β

, so this refinement yields

α \in [3, 4], q_{4} = \frac{4 - α}{β - 4} < \frac{1}{4} .

(7)

For the shift

5

, the original Gershgorin discs only give

∣ α - 5∣ \leq 4

and

∣ β - 5∣ \geq 3

; those bounds do not decide which is closer. The refined interval enclosures do:

1 < 5 - α < 2, 3 < β - 5 < 4, ∣ γ - 5∣ > 8

(8)

for the negative eigenvalues. Therefore the answer for this particular matrix is yes:

μ = 5 : λ_{*} = α \in [3, 4], q_{5} = \frac{5 - α}{β - 5} < \frac{2}{3} .

(9)

Both shifts select the same eigenvector line and alternate signs because

α < μ

. The factor bounds concern asymptotic directional convergence, rather than a universal error reduction from the first iterate for a nonsymmetric matrix. The wider interval