Positive definiteness of $K$ makes the vectors $\phi(x_1),\ldots,\phi(x_n)$ linearly independent. In particular $\Sigma\ne0$, so the maximum Rayleigh quotient is positive and $\lambda>0$. The eigenvector equation gives
$$
v=\frac1{n\lambda}\sum_{i=1}^n
\langle\phi(x_i),v\rangle\phi(x_i),
$$
which lies in their span. Hence $v=\sum_i\alpha_i\phi(x_i)$ for some $\alpha\in\mathbb R^n$.

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Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 218 / 4 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-218/4/d/solution