First off, it turns out that the eigenvalues of are exactly the complex conjugates of those of (the same, if we’re working over . Actually, this isn’t even special to normal operators. Indeed, if has a nontrivial kernel, then we can take the adjoint to find that must have a nontrivial kernel as well. But if our transformation is normal, it turns out that not only do we have conjugate eigenvalues, they correspond to the same eigenvectors as well!
To see this, we do almost the same thing as before. But we get more than just a nontrivial kernel this time. Given an eigenvector we know that , and so it must have length zero. But if is normal then so is :
and so acting by gives the same length as acting by . That is:
thus by the definiteness of length, we know that . That is, is also an eigenvector of , with eigenvalue .
Then as a corollary we can find that not only are the eigenvectors corresponding to distinct eigenvalues linearly independent, they are actually orthogonal! Indeed, if and are eigenvectors of with distinct eigenvalues and , respectively, then we find
Since we must conclude that , and that the two eigenvectors are orthogonal.