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Part BCSIR NET December 2025a-nonzero-real-zero-of-an-invertible-form-rules-out-both-definite-cases-at-once

A nonzero real zero of an invertible form rules out both definite cases at once

Consider the quadratic form f(x,y,z) = [x y z] A [x;y;z], where A is an invertible 3×3 symmetric matrix over Q\mathbb{Q}. Assume that there exists (α,β,γ)∈C3(\alpha,\beta,\gamma) \in \mathbb{C}^{3}{(0,0,0)} such that f(α,β,γ)=0f(\alpha,\beta,\gamma) = 0. Which of the following statements is necessarily true?

  1. A.There exists (a,b,c)∈R3(a,b,c) \in \mathbb{R}^{3}{(0,0,0)} such that f(a,b,c) = 0.
  2. B.If there exists (a,b,c)∈R3(a,b,c) \in \mathbb{R}^{3}{(0,0,0)} such that f(a,b,c) = 0, then (a,b,c)∈Q3(a,b,c) \in \mathbb{Q}^{3}.
  3. C.{(a,b,c)∈Z3(a,b,c) \in \mathbb{Z}^{3} ∣ f(a,b,c) = 0} is a finite set.
  4. D.If there exists (a,b,c)∈Q3(a,b,c) \in \mathbb{Q}^{3}{(0,0,0)} such that f(a,b,c) = 0, then A has a positive eigenvalue and a negative eigenvalue.✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: Every real symmetric matrix is positive definite if det > 0

More on this topic

The chapter behind this: Quadratic forms, signature and definiteness — free to read

From Inner Product Spaces and Forms › Quadratic forms, positive definiteness, Sylvester's law

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