Need verification – Prove a Hermitian matrix $(\textbf{A}^\ast = \textbf{A})$ has only real eigenvalues

Proof: Let eigenvalue $\lambda \neq 0$ such as
$$\textbf{A}\vec{v} = \lambda\vec{v}$$
$$\Rightarrow (\textbf{A}\vec{v})^\ast = (\lambda\vec{v})^\ast$$
$$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast)=(\lambda^\ast\vec{v}^\ast)$$
Right-multiply both sides by $\color{orangered}{\vec{v}}$$$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast \color{orangered}{\vec{v}} )=(\lambda^\ast\vec{v}^\ast \color{orangered}{\vec{v}} )$$
$$\textbf{A}^\ast=\textbf{A}$$
$$\Rightarrow(\vec{v}^\ast\textbf{A}\vec{v})=(\lambda^\ast\vec{v}^\ast\vec{v})$$
$$\Rightarrow(\vec{v}^\ast\lambda\vec{v}) = (\lambda^\ast\vec{v}^\ast\vec{v})$$
$$\Rightarrow(\lambda\vec{v}^\ast\vec{v}) = (\lambda^\ast\vec{v}^\ast\vec{v})$$
$$\Rightarrow \lambda = \lambda^\ast$$
$$\Rightarrow \lambda\in\mathbb{R}$$

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