1
$\begingroup$

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.