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Part CCSIR NET June 2025a-commutator-has-trace-0-so-cayley-hamilton-makes-its-square-scalar

A commutator has trace 0 so cayley hamilton makes its square scalar

Let S,T:C2→C2S, T : \mathbb{C}^{2} \to \mathbb{C}^{2} be C−\mathbb{C}-linear transformations and I denote the identity transformation on C2\mathbb{C}^{2}. Which of the following statements are necessarily true?

  1. A.(ST − TS)2=λI)^{2} = \lambda{}I for some λ∈C\lambda \in \mathbb{C}.✓ Correct
  2. B.The characteristic polynomial of (ST − TS)2)^{2} is (x−λ)2(x - \lambda)^{2} for some λ∈C\lambda \in \mathbb{C}.✓ Correct
  3. C.If ST − TS has only one eigenvalue, then ST − TS =λI= \lambda{}I for some λ∈C\lambda \in \mathbb{C}.
  4. D.If ST − TS has only one eigenvalue, then (ST − TS)2)^{2} is the zero transformation.✓ 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 standard counterexample.

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The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

Related counterexample: Same characteristic polynomial ⇒ similar

More on this topic

The chapter behind this: Characteristic vs minimal polynomial — what each tells you — free to read

From Eigenvalues and Canonical Forms › Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

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