I know, that if $P$ is a projector then $\forall \ k \in \mathbb{N} \rightarrow P^k$ is a projector too. Because $P^2 = P, P^3 = P \times P^2 = P \times P = P, P^4 = P^2 \times P^2 = P ...$. But is the converse true? If, for example, $P^3$ is a projector, does it imply that $P$ is also a projector?
I've tried using the definition of the projector: $$P^3 = P^3 \times P^3 \\ PPPPPP = PPP \\ (PP)(PP)(PP) = PPP$$ I think, I cannot conclude that $(PP) = P$. Also, I've tried finding eigenvalues of $P$: if $P^3$ is a projector then $\det(P^3) = 0$ or $\det(P^3 - I) = \det(P-I)\det(P^2 + P + I) = 0$ so, $P$ has real eigenvalues $0$ or $1$. But this fact also doesn't guarantee that $P$ is a projector, if it isn't diagonalizable, for example.
Could you please giave me any hints for proof, or help to think up a counter example? Thanks a lot for any help!
In finite dimension, every projector is diagonalizable. Then if $P^k$ is a projector, there exists $D, Q$ such that $P^k=QDQ^{-1}$ with $D$ a diagonal matrix.
Since $P^k$ has eigenvalues being $0$ or $1$, then if $k$ is odd, we have $D^{\frac 1 k}=D$ and then $P^k=P$. Hence $P$ is a projector. If $k$ is even, then nothing can be concluded.
Consider for example $P:\mathbb{R} \to \mathbb{R}$, $P(x)=-x$, then $P^2=I$ which is obviously a projector but $P$ is not a projector.