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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 40C / d / Solution

The displayed matrix is an upper bidiagonal matrix. For

λ_{1} = 1

, the right eigenvector equation

(A - I) v = 0

successively gives

v_{2} = v_{3} = \dots = v_{n} = 0.

We may therefore take the unit right eigenvector to be

v = e_{1}

.

For a left eigenvector,

(A^{T} - I) u = 0

gives

u_{i - 1} + (λ_{i} - 1) u_{i} = 0 (2 \leq i \leq n) .

Since

λ_{i} - 1 = - 1/ i

,

u_{i} = i u_{i - 1} .

Choosing the initial scale

u_{1} = 1

produces

u = (1, 2!, 3!, \dots, n!)^{T} .

After normalization,

u = \frac{( 1 , 2 ! , 3 ! , \dots , n ! ) ^{T}}{1 + ( 2 ! ) ^{2} + \dots + ( n ! ) ^{2}} .

The eigenvalue sensitivity is therefore

s (1) = \frac{1}{∣ u ^{T} v ∣} = 1 + (2!)^{2} + \dots + (n!)^{2} \geq n!,

where

n!

is the factorial. Hence

s (λ_{1}) \geq n! .

Solved by gpt-5.6-sol high.

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