Part CCSIR NET December 2024matching-coefficients-gives-c3-equals-c2-either-way

Matching coefficients gives c3 equals c2 either way

Let (V, ⟨ , ⟩) be an inner product space over $\mathbb{C}$ and T : V → V be $a \mathbb{C}-$ linear transformation. Let $v_{1}$ and $v_{2}$ be non-zero vectors in V such that Tv $_{1} = c_{1}v_{1}$ and Tv $_{2} = c_{2}v_{2}$ for some $c_{1}, c_{2} \in \mathbb{C}$ . Let $v_{3} = v_{2} - ($ ⟨ $v_{2}, v_{1}$ ⟩/‖ $v_{1}$ ‖ $^{2})v_{1}$ . Suppose that $v_{3}$ is non-zero and Tv $_{3} = c_{3}v_{3}$ for some $c_{3} \in \mathbb{C}$ . Which of the following statements are true?

A.The set { $v_{1}, v_{2}$ } is linearly independent.✓ Correct
B.If $c_{1} \ne c_{2}$ , then ⟨ $v_{1}, v_{2}$ ⟩ = 0.✓ Correct
C.If $c_{1} = c_{2}$ , then $c_{3} = c_{2}$ .✓ Correct
D.If $c_{1} \ne c_{2}$ , then $c_{3} = c_{2}$ .✓ 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 execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

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Matching coefficients gives c3 equals c2 either way

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