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Part CCSIR NET June 2024reaching-a-positive-value-is-not-definiteness

Reaching a positive value is not definiteness

Let q1(x1,x2)q_{1}(x_{1}, x_{2}) and q2(y1,y2)q_{2}(y_{1}, y_{2}) be real quadratic forms such that there exist (u1,u2),(v1,v2)∈R2(u_{1}, u_{2}), (v_{1}, v_{2}) \in \mathbb{R}^{2} such that q1(u1,u2)=1=q2(v1,v2)q_{1}(u_{1}, u_{2}) = 1 = q_{2}(v_{1}, v_{2}). Define q(x1,x2,y1,y2)=q1(x1,x2)−q2(y1,y2)q(x_{1}, x_{2}, y_{1}, y_{2}) = q_{1}(x_{1}, x_{2}) - q_{2}(y_{1}, y_{2}). Which of the following statements are necessarily true?

  1. A.q is a quadratic form in x1,x2,y1,y2x_{1}, x_{2}, y_{1}, y_{2}✓ Correct
  2. B.There exists (t1,t2)∈R2(t_{1}, t_{2}) \in \mathbb{R}^{2} such that q1(t1,t2)=5q_{1}(t_{1}, t_{2}) = 5✓ Correct
  3. C.There does not exist (s1,s2)∈R2(s_{1}, s_{2}) \in \mathbb{R}^{2} such that q2(s1,s2)=−5q_{2}(s_{1}, s_{2}) = -5
  4. D.Given α∈R\alpha \in \mathbb{R}, there exists a vector w∈R4w \in \mathbb{R}^{4} such that q(w)=αq(w) = \alpha✓ Correct

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 converse assumed.

See pricing

50 are analysed free — try those first.

The trap it tests

Converse assumed

The theorem runs one way. You used it in the other.

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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