Let's say we have a self-adjoint operator acting on an inner product space (real or complex), represented, of course, by a self-adjoint matrix.

I'm looking at the proof for spectral theorem in which you build up a basis out of eigenvectors relying on the fact that the characteristic polynomial will always have roots, both over a real and over a complex field, because eigenvalues of a self-adjoint operator are real.

But what I do not understand is, why do all eigenvalues must necessarily be distinct? How do we conclude that?

After all, spectral theorem says that every self-adjoint operator is always diagonalizable and I know that for a matrix of order $n$ to be diagonalizable, it has to have $n$ distinct eigenvalues.

So, what am I missing here?

Edit: A matrix doesn't have to have n distinct eigenvalues in order to be diagonalizable, but if it does have n distinct eigenvalues it is diagonalizable, guess I was too sloppy and tired to notice such a silly mistake! But I'm leaving the question here ^_^


The identity matrix is self-adjoint and all of its eigenvalues are equal (to $1$). The problem is in your false understanding that for a matrix to be diagonalizable, it has to have $n$ distinct eigenvalues ($n$ being the relevant dimension). That is in incorrect, as the identity matrix (and many others) show. You are probably confused with the true statement that if an $n\times n$ matrix has $n$ distinct eigenvalues, then it is diagonalizable.

  • $\begingroup$ Aaaah, yes, you're right, what was I thinking, the statement only goes one way, of course! I'll have to think twice before asking silly questions here, but I'll leave it here, it can't do any harm :) Thanks! $\endgroup$ – PhysSE is Cancer Jul 15 '13 at 7:33

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.