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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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    $\begingroup$ You need the vector to be non-zero to be able to speak of a eigenvalue. $\endgroup$ – xavierm02 Aug 8 '13 at 17:39
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The only thing missing is to note that, since $v$ is non-zero, you also have $v^*v\neq 0$, therefore you can conclude that $\lambda$ is real.

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