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

A.(ST − TS $)^{2} = λ I$ for some $λ \in C$ .✓
B.The characteristic polynomial of (ST − TS $)^{2}$ is $(x - λ)^{2}$ for some $λ \in C$ .✓
C.If ST − TS has only one eigenvalue, then ST − TS $= λ I$ for some $λ \in C$ .
D.If ST − TS has only one eigenvalue, then (ST − TS $)^{2}$ is the zero transformation.✓

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.

See pricing

50 are analysed free — try those first.

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

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

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