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Part CCSIR NET December 2023nilpotent-index-from-nullity

Nilpotent index from nullity

Let T:R5→R5T : \mathbb{R}^{5} \to \mathbb{R}^{5} be an R−\mathbb{R}-linear transformation. Suppose that (1, −1, 2, 4, 0), (4, 6, 1, 6, 0) and (5, 5, 3, 9, 0) span the null space of T. Which of the following statements are true?

  1. A.The rank of T is equal to 2.✓ Correct
  2. B.Suppose that for every vector v∈R5v \in \mathbb{R}^{5} there exists n such that Tnv=0T^{n}v = 0. Then T2T^{2} must be zero.
  3. C.Suppose that for every vector v∈R5v \in \mathbb{R}^{5} there exists n such that Tnv=0T^{n}v = 0. Then T3T^{3} must be zero.✓ Correct
  4. D.(−2, −8, 3, 2, 0) is contained in the null space of T.✓ 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 invariants don't determine.

See pricing

50 are analysed free — try those first.

The trap it tests

Invariants don't determine

Two objects sharing an invariant were treated as the same object.

Drill statements like this

Related counterexample: Equivalent matrices are similar

More on this topic

The chapter behind this: Linear transformations, matrices and similarity — free to read

From Vector Spaces and Linear Maps › Linear transformations, matrix representation, change of basis

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