Prove that a matrix $\mathbf{A} \in M_n$ is similar to a Hermitian matrix if and only if it is diagonalizable and has real eigenvalues

Question

So if the matrix $\mathbf{B}$ is Hermitian, that also means its diagonalizable. And if $\mathbf{A}$ is similar to $\mathbf{B}$ then there exists an invertible matrix $\mathbf{S}$ such that $\mathbf{A}=\mathbf{S}-1\mathbf{B}\mathbf{S}$. Using that equation and the equation for $\mathbf{A}$'s eigenvalues, we can prove that $\mathbf{A}$ is also diagonalizable and has real eigenvalues. I'm confused though, on how to prove this statement the other way. How do we prove two matrices are similar with the given information?

Answer 1

If $A$ is diagonalizable and it has real eigenvalues, then it is similar to a diagonal matrix $D$ such that all the entries of the main diagonal are real. But then $D$ is Hermitian.