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

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

50 are analysed free — try those first.

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