10E (Linear Algebra)

Solution

χ_{A} (t) = det (t I - A)

, then

χ_{A} (A) = 0

For the proof, the adjugate identity gives

(t I - A) adj (t I - A) = χ_{A} (t) I .

Write

adj (t I - A) = B_{0} + B_{1} t + \dots + B_{n - 1} t^{n - 1}

and compare coefficients. Multiplying the resulting recurrence by successive powers of

A

and summing telescopically yields

χ_{A} (A) = 0

p = q χ_{A} + r

with

de g r < n

. Therefore

p (A) = r (A)

by Cayley-Hamilton, while

p (λ) = r (λ)

for every eigenvalue

λ

, since

χ_{A} (λ) = 0

For the displayed matrix,

χ_{A} (t) = (t - 2) (t^{2} + 1) .

t^{1000}

modulo this polynomial has the form

a t^{2} + c

, because its values at

i

and

- i

are both

1

- a + c = 1

and

4 a + c = 2^{1000}

give

r (t) = \frac{2 ^{1000} - 1}{5} t^{2} + \frac{2 ^{1000} + 4}{5} .

Since

(A^{2})_{11} = 3

, the requested entry is

(A^{1000})_{11} = \frac{4 \cdot 2 ^{1000} + 1}{5} .

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