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Part CCSIR NET December 2024positive-definite-says-nothing-about-this-basis

Positive definite says nothing about this basis

Let V be a two-dimensional real vector space with a basis {v1,v2v_{1}, v_{2}}. Let f:V×V→Rf : V \times V \to \mathbb{R} be a symmetric bilinear form. Let r=f(v1,v1),s=f(v1,v2),t=f(v2,v2)r = f(v_{1}, v_{1}), s = f(v_{1}, v_{2}), t = f(v_{2}, v_{2}). Which of the following statements are necessarily true?

  1. A.f(v2,v1)=−sf(v_{2}, v_{1}) = -s.
  2. B.If r = s = t, then either f(v, v) ≥ 0 for all v ∈ V or f(v, v) ≤ 0 for all v ∈ V.✓ Correct
  3. C.If f is positive definite, then r ≠ s, s ≠ t and r ≠ t.
  4. D.If f is positive definite, then f(v1,v2)=0f(v_{1}, v_{2}) = 0.

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, and why this one tests hypothesis dropped.

See pricing

50 are analysed free — try those first.

The trap it tests

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

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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