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Part CCSIR NET June 2025q-of-v-vanishing-makes-the-quadratic-in-alpha-linear-so-there-is-one-root-not-two

Q of v vanishing makes the quadratic in alpha linear so there is one root not two

Let B(v, w) be a nondegenerate symmetric bilinear form on and let q(v) = B(v, v) be the corresponding quadratic form. Suppose there exist vectors such that B(v, v) = 0 and B(v, w) ≠ 0. Which of the following statements are necessarily true?

  1. A.B(w, w) = 0
  2. B.There exists an such that .
  3. C.There are infinitely many such that .
  4. D.q is equivalent to the quadratic form for all .

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 boundary and endpoint.

See pricing

50 are analysed free — try those first.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

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 FormsQuadratic forms, positive definiteness, Sylvester's law

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