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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-339/2/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 339 / 2 / c / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

Write a test vector as

(u, v)

and put

s = \sum_{i} u_{i}

,

t = \sum_{j} v_{j}

. Its quadratic form against the block-constant matrix is

a (s^{2} + t^{2}) + 2 b s t .

Since

a \geq ∣ b ∣

, both

a

and

a - ∣ b ∣

are nonnegative, and

a (s^{2} + t^{2}) + 2 b s t \geq a (s^{2} + t^{2}) - 2∣ b ∣∣ s t ∣ \geq (a - ∣ b ∣) (s^{2} + t^{2}) \geq 0.

Equivalently it is

\frac{a + b}{2} (s + t)^{2} + \frac{a - b}{2} (s - t)^{2} .

Therefore the bipartite block-constant positive semidefinite matrix satisfies

(a J_{n, n} b J_{m, n} b J_{n, m} a J_{m, m}) ⪰ 0

.

For positive block sizes, the condition is also necessary: choose

s = t

and

s = - t

to obtain

a + b \geq 0

and

a - b \geq 0

. Equality is allowed, and vectors whose sums vanish in both blocks lie in the kernel. Positive semidefiniteness, rather than positive definiteness, is the conclusion.

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