# Complex Endomorphism $f$ with orthogonal Eigenspaces implies that $f$ is unitary

I already posted this question earlier today without much success. After 3 hours and only 10 views I decided to give it another shot :) Of course I'll delete the second post!

Let $$V \ne 0$$ be an finite-dimensional inner complex vectorspace and $$f$$ an endomorphism on said space such that there exists a $$m \in \mathbb{N}_{\ge 1}$$ with $$f^m=id_V$$. Given that all eigenspaces are orthogonal, I have to show that $$f$$ is unitary. The converse I already have proven.

I thought about starting with $$f^m=id_V$$:

If we take an eigenvector $$x$$ to the eigenvalue $$λ$$ we get: $$f(x)=λx$$, therefore $$f^m(x)=λ^mx$$. But $$f^m(x)=x$$ since $$f^m=id_V$$. Together: $$λ^mx=x$$, so: $$λ^m=1$$.

It follows that all eigenvalues of $$f$$ are $$m$$-th roots of unity and then: $$\bar{\lambda}=1/λ$$

This is the part where I unfortunately get stuck on :P

Any Ideas?

~Cedric :)