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A 3X3 matrix P is such that, P^3 = P. Then the eigenvalues of P are:

1) 1, 1, -1

2) 1, 0.5 + j0.866, 0.5 - j0.866

3) 1, -0.5 + j0.866, -0.5 - j0.866

4) 0, 1, -1

Suppose $x$ is an eigenvector of $P$, and let $\lambda$ be corresponding eigenvalue, then

$$Px=\lambda x$$

Also

$$P^3x=\lambda^3 x$$.

We are given $P^3=P$, which means

$$P^3x=Px \Rightarrow \lambda^3 x = \lambda x \Rightarrow (\lambda^3-\lambda)x=0$$

Since $x$ is non-zero, $\lambda^3-\lambda = 0$. Solving it, we get $\lambda={0,1,-1}$
option 4??

Yes but how?
every square matrix saisfies its own characterstic equation...therefore,

λ^3-λ=0

λ=0,1,-1..

but since i am deriving characterstic equation back from given matrix condition, i am not sure about it.
then why not option 1 ???
@shubhanshu because roots of λ^3-λ=0 will be 0,1,-1...

1
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