# Eigenvectors of a symmetrical matrix

Accepting the fact that, as 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 could I prove that a symmetrical matrix (real hermitian) has a set of N orthonormal real eigenvectors?

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.