Prove that the eigenvalues of a real symmetric matrix are real I am having a difficult time with the following question. Any help will be much appreciated.

Let $A$ be an $n×n$ real matrix such that $A^T = A$. We call such matrices “symmetric.” Prove that the eigenvalues of a real symmetric matrix are real (i.e. if $\lambda$ is an eigenvalue of $A$, show that $\lambda = \overline{\lambda}$ )


Consider the real operator $$u := (x \mapsto Ax)$$ for all $x \in \mathbb{R}^{n}$  and the complex operator $$\tilde{u} := (x \mapsto Ax) $$ for all $x \in \mathbb{C}^{n}$. Both operators have the same characteristic polynomial, say $p(\lambda) = \det(A - \lambda I)$. Since $A$ is symmetric, $\tilde{u}$ is an hermitian operator. For the spectral theorem for hermitian operators all the eigenvalues (i.e. the roots of the $p(\lambda)$) of $\tilde{u}$ are real. Hence, all the eigenvalues (i.e. the roots of the $p(\lambda)$) of $u$ are real.
We have shown that the eigenvalues of a symmetric matrix are real numbers as a consequence of the fact that the eigenvalues of an Hermitian matrix are reals.
