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Question: If A<sup>k</sup> = 0, for some value of k, (I – A)<sup>p</sup> = I + A + A<sup>2</sup> + …. A<sup>k–...

If Ak = 0, for some value of k, (I – A)p = I + A + A2 + …. Ak–1, thus p is (A is nilpotent with index k).

A

–1

B

–2

C

–3

D

None of these

Answer

–1

Explanation

Solution

Let B = I + A + A2 + … + Ak–1

Post multiply both sides by (I – A), so that

B(I – A) = (I + A + A2 + … + Ak–1) (I – A)

= I – A + A – A2 + A2 – A3 + … –Ak–1 + Ak–1 – Ak

= I – Ak = I, since Ak = 0

Ž B = (I – A)–1

Hence (I – A)–1 = I + A + A2 + … + Ak–1.

Thus, p = –1.

Hence (1) is correct answer.