Eigenvalues of real, symmetric matrices are real. But what can we say about their eigenvectors?
The question arose as I was trying to comprehend the proof of the Real Spectral theorem given in "Linear Algebra - A Modern Introduction" by D.Poole.
I quote:
I could not see why and tried to work a proof myself.The Spectral Theorem
Let A be diagonalizable n x n real matrix. Then A is symmetric if and only if it is orthogonally diagonalizable
Proof: We have already proved the "if" part as T. 5.17. To prove the "only if" part implication we proceed by induction on n. For n=1 there is nothing to
do since a 1 x 1 matrix is already in diagonal form. Now assume that every k x k real symmetric matrix with real eigenvalues is orthogonally diagonalizable.
Let n = k + 1 and let A be an n x n real symmetric matrix with real eigenvalues.
Let be one of the eigenvalues of A and let be a corresponding eigenvector. Then is a real vector (why?) ...... (and the proof goes on)
For example if is an eigenvalue of A and x is an eigenvector of A, corresponding to the eigenvalue, I think the following is true.
. Taking complex conjugates (please excuse me if I apply conjugation wrong - complex numbers still look very alien to me).
which implies that the complex conjugate of x is eigenvector, associated with the same eigenvalue. It gets even better
which implies that is an eigenvector to the same eigenvalue too. From here I was hoping to find a way to proof that x is equal to its complex conjugate which will conclude
the proof. I could not see it...
So... I ran to this great forum, but checked quite non thoroughly some posts in similar-themed sites and forums on the way. They seem to indicate that if we consider A to be in then in the general case of real symmetric matrix A the eigenvectors may be real or complex, which seems to indicate that the conclusion that eigenvectors are real may be coming from the proof of the theorem itself. But to that point (see the quote) we did almost nothing (merely an induction step).
Well, help?