Part CCSIR NET June 2025a-vector-killed-by-a-transpose-confines-the-whole-image-to-a-hyperplane

A vector killed by a transpose confines the whole image to a hyperplane

Let $M_{2} (R)$ denote the space of real 2 × 2 matrices. Let S be the vector subspace of $M_{2} (R)$ comprising of all symmetric matrices. Let $F : M_{2} (R) \to S$ be the map defined by F(X) = XXᵀ. Let DF $_{A} : M_{2} (R) \to S$ be the derivative of F at $A \in M_{2} (R)$ . Which of the following statements are true?

A.If AAᵀ = I, then DF $_{A} : M_{2} (R) \to S$ is surjective.✓
B.If AAᵀ = I, then DF $_{A} : M_{2} (R) \to S$ need not be surjective.
C.If A is invertible, then DF $_{A} : M_{2} (R) \to S$ is surjective.✓
D.If A is not invertible, then DF $_{A} : M_{2} (R) \to S$ is surjective.

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: Equivalent matrices are similar

A vector killed by a transpose confines the whole image to a hyperplane

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