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Part CCSIR NET December 2025non-negative-quietly-admits-zero-which-breaks-the-strict-positivity-the-diagonalizability-argument-actually-needs

Non negative quietly admits zero which breaks the strict positivity the diagonalizability argument actually needs

Let V be a finite-dimensional R−\mathbb{R}-vector space and T:V→V a linear operator such that T2T^{2} is diagonalizable over R\mathbb{R}. Which of the following statements are necessarily true?

  1. A.If T is not diagonalizable over R\mathbb{R}, then T2T^{2} has an eigenvalue ≤0.✓
  2. B.If T2T^{2} has only negative eigenvalues, then dim V is an even integer.✓
  3. C.If T2T^{2} has only non-negative eigenvalues, then T is diagonalizable over R\mathbb{R}.
  4. D.For each non-zero v∈V, {v,Tv,T2vT^{2}v} is linearly dependent.

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.

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50 are analysed free — try those first.

Related counterexample: Diagonalisable ⇒ invertible

More on this topic

The chapter behind this: Diagonalisability — the criteria card — free to read

From Eigenvalues and Canonical Forms › Diagonalisability criteria

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