Madeeasy workbook

Question

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

• A. $1,1,1$
• B. $1,0.5+j(0.886),0.5-j(0.866)$
• C. $1,-0.5+j(0.866),-0.5-j(0.886)$
• D. $0,1,-1$

Answer 1

Apply The Cayley-Hamilton Theorem

Answer 2

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