For a real symmetric $3 \times 3$ matrix $A$ having eigenvalues $\lambda_1 = 2$, $\lambda_2 = 5$, and $\lambda_3 = 9$, the maximum possible value of the Rayleigh quotient $R(\vec{x}) = \frac{\vec{x}^T A \vec{x}}{\vec{x}^T \vec{x}}$ for any non-zero vector $\vec{x} \in \mathbb{R}^3$ is:

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### 1. Strategy & Theoretical Principles By the Rayleigh-Ritz theorem for real symmetric matrices: $$\lambda_{\min} \le R(\vec{x}) = \frac{\vec{x}^T A \vec{x}}{\vec{x}^T \vec{x}} \le \lambda_{\max}$$ where: - The minimum value of the Rayleigh quotient is the smallest eigenvalue $\lambda_{\min}$ - The maximum value of the Rayleigh quotient is the largest eigenvalue $\lambda_{\max}$ - Equality is attained when $\vec{x}$ is an eigenvector corresponding to $\lambda_{\min}$ or $\lambda_{\max}$. ### 2. Step-by-Step Calculation The given eigenvalues of $A$ are $\{2, 5, 9\}$. Therefore: $$\lambda_{\max} = 9$$ $$\max_{\vec{x} \neq 0} R(\vec{x}) = 9.0$$ ### 3. Distractor Forensics - Candidate dividing by dimension (9/3 = 3) or calculating trace (16). ### 4. Exam Trap & Standard Reference - **Trap Alert:** For real symmetric matrices, the maximum of the quadratic form on the unit sphere is always the maximum eigenvalue $\lambda_{\max}$. - **Standard Reference:** Strang, G., *Linear Algebra and Its Applications*, 4th Edition, Chapter 6.

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