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If $\lambda$ is the largest eigenvalue of a real symmetric $n \times n$ matrix $H$, how can I show that: $$\forall v \in \mathbb{R^n}, ||v||=1 \implies v^tHv\leq \lambda$$

Thank you.

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    http://math.stackexchange.com/questions/9302/norm-of-a-symmetric-matrix2011-03-29

3 Answers 3