Proof Hermitian matrix has only real eigenvalues

1 Hermitian matrix

If matrix \(A\) satisfies following \[ A = A^{*} \]
then \(A\) is called Hermitian matrix. \(A^{*} \Rightarrow\) Transpose and Conjugation.
\( A =\begin{bmatrix} 1 & 2 & 1+2i \\ 4 & 5 & 6 \\ 7 & 8 & 10 \end{bmatrix} \Rightarrow A^{*} = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 1-2i & 6 & 10 \end{bmatrix} \)

let \(\vec{v}\) is a eigenvalue of \(A\) and \(\vec{v} \neq 0\)
we have
\begin{align*} A^{*} \vec{v} &= \lambda \vec{v} \\ \left( A^{*} \vec{v} \right)^{*} &= \left( \lambda \vec{v} \right)^{*} \\ \vec{v}^{*} A &= \vec{v}^{*} \lambda^{*} \\ \vec{v}^{*} A \vec{v} &= \vec{v}^{*} \lambda^{*} \vec{v} \quad \mbox{ multiply both side with } \vec{v} \\ \vec{v}^{*} \lambda \vec{v} &= \vec{v}^{*} \lambda^{*} \vec{v} \quad \mbox{ where } A \vec{v} = \lambda \vec{v} \\ \lambda &= \lambda^{*} \end{align*}
\(\Rightarrow \lambda\) is \(\mathbf{R}\)