Operator norm and eigenvalue inequality

Question

Can I say that $\|A\| < s$ where $A \in \mathbb{R}^{3 \times 3}$ is a symmetric, positive definite matrix and $s$ is the maximum eigenvalue of $A$. Here the norm used is operator norm.

Answer 1

If $A$ is symmetric and positive definite, then its eigenvalues are real and positive. The spectral norm of $A$ is given by

$$\|A\|_2 := \sigma_{\max} (A) = \sqrt{\lambda_{\max} (A^T A)} = \sqrt{\lambda_{\max} (A^2)} = \sqrt{\lambda_{\max}^2 (A)} = |\lambda_{\max} (A)| = \lambda_{\max} (A)$$