Prove that $\Phi^3$ is an orthogonal projection

Question

Let V be a n-dimensional unitary vector space and $\Phi:V\to V$ an endomorphism such that $\Phi^2 = \Phi^*$. Prove that $\Phi^3$ is an orthogonal projection.

I'm afraid I'm lacking any decent starting point here. I know that a projection is orthogonal iff it is self-adjoint but that didn't get me anywhere. I also noticed that $\Phi^3$ is normal since $$\Phi \Phi^* = \Phi \Phi^2 = \Phi^2 \Phi = \Phi^* \Phi$$

I would greatly appreciate some clarification. How can we prove this?

Thank you.

Answer 1

$\Psi$ is an orthogonal projection iff $\Psi^*=\Psi$ and $\Psi^2=\Psi$. Let $\Psi=\Phi^3$, then $\Psi=\Phi\Phi^2=\Phi\Phi^*$ and $\Psi=\Phi^2\Phi=\Phi^*\Phi$. It is easy to see that $\Psi^*=\Psi$. We have also $$ \Psi^2=(\Phi^*\Phi)(\Phi\Phi^*)=\Phi^*\Phi^2\Phi^*=(\Phi^*)^3=(\Phi^3)^*=\Psi^*=\Psi. $$

Answer 2

More general, if $a$ is an element of a $*$-algebra with $a^2=a^*$, then $a^3$ is an orthogonal projection:

$a^2=a^*$ implies $(a^*)^2=a$. Hence, $$(a^3)^* = (a^*)^3=(a^*)^2 a^* = a a^* = a a^2 = a^3,$$ and $$(a^3)^2 = (a^2)^3=(a^*)^3=(a^3)^*=a^3.$$