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$$\vec{v}^{t}\textbf{A}\vec{v} > \textbf{0}\text{ and }\textbf{A}\vec{v} = \lambda\vec{v}\quad \Rightarrow \lambda>\textbf{0}\quad(\mathbb{F}=\mathbb{R}) $$

proof: $$\vec{v}^{t}\textbf{A}\vec{v} > \textbf{0} \text{ and } \textbf{A}\vec{v}=\lambda\vec{v} $$ $$\Rightarrow\vec{v}^{t}\lambda\vec{v} > \textbf{0} $$ $$\Rightarrow\lambda\vec{v}^{t}\vec{v} > \textbf{0} $$ $$\Rightarrow\lambda\langle v, v\rangle> \textbf{0} $$ $$\vec{v} \neq \vec{0}$$ $$\Rightarrow\langle v, v\rangle > \textbf{0} $$ $$\Rightarrow\lambda > \textbf{0}$$

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  • $\begingroup$ Seems right to me.. $\endgroup$ – user91011 Oct 10 '13 at 4:43
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Yes, you are correct. Another perspective. Let $v_{min}$ be the unit norm eigenvector corresponding to $\lambda_{min}$, the lowest eigenvalue. \begin{align} 0<\min_{||v||=1}v^TAv\leq v_{min}^TAv_{min}=\lambda_{min} \end{align} Can you prove the middle inequality is in fact an equality?

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Yes, you are right. Everything is correct.

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