I am studying the spectral theorem for matrices, and the book says that if a $nxn$ matrix A is real and symmetric then its diagonalizable over $\mathbb{R}$. And that this fact is a corollary of the Spectral Theorem for the complex case of normal matrices.
Although I agree that since $A$ is symmetric then $A$ is normal hence it implies that $A$ is diagonalizable over $\mathbb{C}$, and moreover it is easy to prove that all eigenvectors are real. But how can I see that all eigenvectors are also real?
Thanks in advance.
Best Answer
Suppose $v$ is a complex eigenvector of $A$, a real symmetric matrix, with corresponding real eigenvalue $\lambda$. Note that \begin{align} \lambda \overline{v}=\overline{\lambda v} = \overline{Av}= A\overline{v} \end{align} then $\overline{v}$ is also an eigenvector. Hence $v+\overline{v}$ is a real eigenvector. So pick this one.