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Part BCSIR NET December 2024the-span-of-commutators-is-all-of-sl

The span of commutators is all of sl

Let V be the R−\mathbb{R}-vector space of 5 × 5 real matrices. Let S = {AB − BA | A, B ∈ V} and W denote the subspace of V spanned by S. Let T:V→RT : V \to \mathbb{R} be the linear transformation mapping a matrix A to its trace. Which of the following statements is true?

  1. A.W = ker(T)✓ Correct
  2. B.W ⊊ ker(T)
  3. C.W ∩ ker(T) ⊊ W
  4. D.W ∩ ker(T) ⊊ ker(T)

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: AB = I ⇒ BA = I for all matrices

More on this topic

The chapter behind this: Determinants, trace and block matrices — free to read

From Determinants and Matrix Tricks › Determinants, trace, block matrices, rank inequalities

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