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A $3\times 3$ matrix $P$ is such that $P^3 =P$.Then the eigenvalues of $P$ are

1. $1,1,1$
2. $1,0.5+j(0.886),0.5-j(0.866)$
3. $1,-0.5+j(0.866),-0.5-j(0.886)$
4. $0,1,-1$

$P^{3} - P = 0$

$P \Leftrightarrow \lambda$

$\Rightarrow \lambda ^{3} - \lambda = 0$

$\Rightarrow \lambda (\lambda ^{2}-1) = 0$

$\Rightarrow \lambda (\lambda -1)(\lambda + 1) = 0$

$\fn_phv \Rightarrow \lambda = 0,1,-1 ( Answer )$

option D

given P3=P

=> P(P2-I)=0

=> |P|=0   and   |P2-I|=0 (if multiplication of 2 matrix is 0,then determinant of individual 2 matrixes are 0)

|P|=0 => so one eigen value of P is 0.

|P2-I|=0  => so one eigen value of P2 is 1.We know that if λ is eigen value of P,then λ2 is eigen value is P2.Here λ2=1.So λ=1 and λ = -1.

so 3 eigen values are 0,1,-1