If T is hermitian or skew-hermitian, then two distinct eigenvalues impliy that two eigenvectors are orthogonal.

But, there is another theorem that if T is hermitian or skew hermitian , then there exists eigenvectors foming a orthonormal basis.

Second theorem does not specify about eigenvalues. Am I under assumption that eigenvalues are distinct for second theorem?

If not, what is so special about first theorem? Orthonormality implies orthogonal. If eigenvalues could be same for second theorem, then what does first theorem tell?


Suppose $T$ is Hermitian. The second theorem then says there is some orthonormal basis. It does not concern itself abut whether eigenvalues might be equal. And it does not say that every eigenvector of $T$ is proportional to one of the members of that otrthonormal basis. And in fact, if two eigenvalues $e_1, e_2$ of $T$ happen to be equal, then any linear combination $a{\bf v_1} + b{\bf v_2} $ is also an eigenvector (sharing that same eigenvalue).

What the first theorem says is that this is the only case in which you can get two non-orthogonal eigenvectors for a hermition (or skew-hermitian) matrix or operator.

  • $\begingroup$ I dont understand your second paragraph. "It is the only case you can get non-orthogonal." $\endgroup$ – Kein Nov 5 '14 at 23:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.