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Part CCSIR NET December 2025the-anticommuting-space-allows-b21-to-be-nilpotent-which-is-never-diagonalisable

The anticommuting space allows b21 to be nilpotent which is never diagonalisable

Let A∈M4(C)A \in M_{4}(\mathbb{C}) be such that A2=IA^{2} = I and Trace(A) = 0. Which of the following statements are necessarily true?

  1. A.The set {B∈M4(C)B \in M_{4}(\mathbb{C}) ∣ AB+BA=0} is an 8-dimensional C−\mathbb{C}-vector space.✓ Correct
  2. B.The set {B∈M4(C)B \in M_{4}(\mathbb{C}) ∣ AB−BA=0} is an 8-dimensional C−\mathbb{C}-vector space.✓ Correct
  3. C.There exists B∈M4(C)B \in M_{4}(\mathbb{C}) such that AB+BA=0,B2=I=0, B^{2}=I and Trace(B)=0.✓ Correct
  4. D.If B∈M4(C)B \in M_{4}(\mathbb{C}) is such that AB+BA=0, then B is diagonalisable.

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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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

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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