Solution:
Since A is idempotent, we have
A² = A
(I – A) ² = I² + A² – 2IA
(I – A) ² = I + A – 2A
(I – A) ² = I -A
》 I – A is idempotent
Since A is inevitable, A‐¹ exists
A² = A
Multiply both sides by A-¹
A-¹(AA) = A-¹A
(A-¹A) A = A-¹A
IA = I
》 A= I
Let λ be the given value of A corresponding to the vector.
Then, Av = λv
》 A³v = λv
》 A(Av) = λv
》 A(λv) = λv
》λ (Av) = λv
》λ (λv) = λv
》λ²v = λv
》λ²v – λv = 0
》 (λ²-λ) v = 0
But v ≠ 0
》λ² – λ = 0
》λ (λ-1) = 0
》λ = 0, 1
჻ The only eigenvalues of A are 0 and 1