 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-1/31l/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

For a positive semidefinite matrix, the sum of the eigenvalues is its trace. Thus

S = {M = M^{T} : M ⪰ 0, tr M \leq s} .

M, N \in S

and

0 \leq t \leq 1

, then

tM + (1 - t) N ⪰ 0

and

tr (tM + (1 - t) N) = t tr M + (1 - t) tr N \leq s .

Hence $S$ is convex.

 Read the full article

 New to topics? Read the docs here!