Let be a matrix in .
Prove that if ia an eigenvector with eigenvalue and if , then .
Perhaps I'm misunderstanding something here, but isn't just the inner (scalar) product of the vector with itself, also known as ?
If so then it's obvious it can't be zero, and this much is true for ANY eigenvector of ANY operator/matrix, since , and zero cannot be an eigenvector...
Yeah, I totally missed the complex thing: I'm used to call a complex matrix unitary...but then you have a counterexample: the unity matrix is (special) orthogonal and all its eigenvectors are 1, and here up you just showed one of its eigenvectors for which ...! Thus, the claim is false
Let be eigenvectors for the three eigenvalues. Then and so (because P is orthogonal). Thus . Since , it follows that . Similarly .
If in addition then for all (because is a basis for ). In particular, , where is the vector whose coordinates are the complex conjugates of those of X. But that implies that : contradiction. Therefore .