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Accepting the fact that, like all others normal matrices, a hermitian matrix (with $N$ lines) has a set of $N$ orthonormal complex eigenvectors (with real eigenvalues, degenerate or not), how do I prove that a symmetric matrix (real hermitian) has a set of $N$ orthonormal real eigenvectors?

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It follows simply from the fact that the eigenvalues of the symmetric matrix are real. Assume that $A$ has a real eigenvalue $\lambda$ and a nonzero eigenvector $u=v+iw$, where $v$ and $w$ are real. Then it follows $Au=\lambda u$ that $Av=\lambda v$ and $Aw=\lambda w$ and hence $v$ and $w$ are in the same eigenspace associated with $\lambda$. Consequently, the eigenvectors can be chosen to be real.

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  • $\begingroup$ I see... But, let's say, we have an eigenvalue lamba with L orthonormal eigenvectors (L <= N, of course). It's not obvious to me that from the imaginary and real parts of each vector (2L real vectors) I could get L real and linearly independent vectors. $\endgroup$ – Rafael Oct 4 '13 at 11:41

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