Substituting $v=\sum_j\alpha_j\phi(x_j)$ into $\Sigma v=\lambda v$ gives
$$
\frac1n\sum_i(K\alpha)_i\phi(x_i)
=\lambda\sum_i\alpha_i\phi(x_i).
$$
Linear independence yields $K\alpha=n\lambda\alpha$. The unit-norm constraint gives
$$
1=\langle v,v\rangle=\alpha^\top K\alpha.
$$

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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